String is used uninitialized [duplicate]



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This question already has an answer here:



  • Update (int) variable in C inside a function [duplicate]

    5 answers



  • Change string with malloc in other function

    2 answers



  • Passing Char pointer in C function [duplicate]

    1 answer



  • How do I modify a pointer that has been passed into a function in C?

    5 answers



I have created this very simple program. My goal is to have the output say String: hello world James but I want the hello world to be malloced in my test_function. Can someone explain to me how I can make my_intro = "hello world" and name my_name = "James". This problem is revolves around how I can parse the malloc-ed char value back to my main function. This is not a duplicate of changing ints. This is parsing char *



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(my_intro, my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

a_intro = intro;
a_name = name;




But the error I get is:



testcode.c: In function 'main':
testcode.c:10:5: error: 'my_intro' is used uninitialized in this function [-Werr
or=uninitialized]
test_function(my_intro, my_name);
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
testcode.c:10:5: error: 'my_name' is used uninitialized in this function [-Werro
r=uninitialized]
cc1.exe: all warnings being treated as errors


Another possible solution which doesn't work:



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(&my_intro, &my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

*a_intro = intro;
*a_name = name;




Error:



testcode.c: In function 'main':
testcode.c:10:19: error: passing argument 1 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c:10:30: error: passing argument 2 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c: In function 'test_function':
testcode.c:22:14: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_intro = intro;
^
testcode.c:23:13: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_name = name;
^
cc1.exe: all warnings being treated as errors









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Nov 15 '18 at 9:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • What is unclear about the errors? test_function takes 2 arguments, you call it with 3. There is no variable my_str in main, you only have my_intro and my_name.

    – mch
    Nov 15 '18 at 9:06






  • 1





    In C all function arguments are passed by value. That means they are copied. The arguments inside a function are distinct and separate variables from what was used when calling the function. Modifying the copy inside the function will not modify the original variable. To solve that (and your error) please do some research about emulating pass by reference in C.

    – Some programmer dude
    Nov 15 '18 at 9:07











  • @mch its not. I tried the solution in that answer already. That answer is about ints. I tried what solution for my char. It didn't work

    – Kiwa
    Nov 15 '18 at 9:12












  • @Kiwa The fundamentals of pass-by-value don't change depending on the type of your variables.

    – melpomene
    Nov 15 '18 at 9:16











  • @melpomene I've edited my question with the logic of the duplicate.

    – Kiwa
    Nov 15 '18 at 9:17

















0
















This question already has an answer here:



  • Update (int) variable in C inside a function [duplicate]

    5 answers



  • Change string with malloc in other function

    2 answers



  • Passing Char pointer in C function [duplicate]

    1 answer



  • How do I modify a pointer that has been passed into a function in C?

    5 answers



I have created this very simple program. My goal is to have the output say String: hello world James but I want the hello world to be malloced in my test_function. Can someone explain to me how I can make my_intro = "hello world" and name my_name = "James". This problem is revolves around how I can parse the malloc-ed char value back to my main function. This is not a duplicate of changing ints. This is parsing char *



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(my_intro, my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

a_intro = intro;
a_name = name;




But the error I get is:



testcode.c: In function 'main':
testcode.c:10:5: error: 'my_intro' is used uninitialized in this function [-Werr
or=uninitialized]
test_function(my_intro, my_name);
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
testcode.c:10:5: error: 'my_name' is used uninitialized in this function [-Werro
r=uninitialized]
cc1.exe: all warnings being treated as errors


Another possible solution which doesn't work:



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(&my_intro, &my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

*a_intro = intro;
*a_name = name;




Error:



testcode.c: In function 'main':
testcode.c:10:19: error: passing argument 1 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c:10:30: error: passing argument 2 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c: In function 'test_function':
testcode.c:22:14: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_intro = intro;
^
testcode.c:23:13: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_name = name;
^
cc1.exe: all warnings being treated as errors









share|improve this question















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Nov 15 '18 at 9:16


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • What is unclear about the errors? test_function takes 2 arguments, you call it with 3. There is no variable my_str in main, you only have my_intro and my_name.

    – mch
    Nov 15 '18 at 9:06






  • 1





    In C all function arguments are passed by value. That means they are copied. The arguments inside a function are distinct and separate variables from what was used when calling the function. Modifying the copy inside the function will not modify the original variable. To solve that (and your error) please do some research about emulating pass by reference in C.

    – Some programmer dude
    Nov 15 '18 at 9:07











  • @mch its not. I tried the solution in that answer already. That answer is about ints. I tried what solution for my char. It didn't work

    – Kiwa
    Nov 15 '18 at 9:12












  • @Kiwa The fundamentals of pass-by-value don't change depending on the type of your variables.

    – melpomene
    Nov 15 '18 at 9:16











  • @melpomene I've edited my question with the logic of the duplicate.

    – Kiwa
    Nov 15 '18 at 9:17













0












0








0









This question already has an answer here:



  • Update (int) variable in C inside a function [duplicate]

    5 answers



  • Change string with malloc in other function

    2 answers



  • Passing Char pointer in C function [duplicate]

    1 answer



  • How do I modify a pointer that has been passed into a function in C?

    5 answers



I have created this very simple program. My goal is to have the output say String: hello world James but I want the hello world to be malloced in my test_function. Can someone explain to me how I can make my_intro = "hello world" and name my_name = "James". This problem is revolves around how I can parse the malloc-ed char value back to my main function. This is not a duplicate of changing ints. This is parsing char *



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(my_intro, my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

a_intro = intro;
a_name = name;




But the error I get is:



testcode.c: In function 'main':
testcode.c:10:5: error: 'my_intro' is used uninitialized in this function [-Werr
or=uninitialized]
test_function(my_intro, my_name);
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
testcode.c:10:5: error: 'my_name' is used uninitialized in this function [-Werro
r=uninitialized]
cc1.exe: all warnings being treated as errors


Another possible solution which doesn't work:



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(&my_intro, &my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

*a_intro = intro;
*a_name = name;




Error:



testcode.c: In function 'main':
testcode.c:10:19: error: passing argument 1 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c:10:30: error: passing argument 2 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c: In function 'test_function':
testcode.c:22:14: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_intro = intro;
^
testcode.c:23:13: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_name = name;
^
cc1.exe: all warnings being treated as errors









share|improve this question

















This question already has an answer here:



  • Update (int) variable in C inside a function [duplicate]

    5 answers



  • Change string with malloc in other function

    2 answers



  • Passing Char pointer in C function [duplicate]

    1 answer



  • How do I modify a pointer that has been passed into a function in C?

    5 answers



I have created this very simple program. My goal is to have the output say String: hello world James but I want the hello world to be malloced in my test_function. Can someone explain to me how I can make my_intro = "hello world" and name my_name = "James". This problem is revolves around how I can parse the malloc-ed char value back to my main function. This is not a duplicate of changing ints. This is parsing char *



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(my_intro, my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

a_intro = intro;
a_name = name;




But the error I get is:



testcode.c: In function 'main':
testcode.c:10:5: error: 'my_intro' is used uninitialized in this function [-Werr
or=uninitialized]
test_function(my_intro, my_name);
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
testcode.c:10:5: error: 'my_name' is used uninitialized in this function [-Werro
r=uninitialized]
cc1.exe: all warnings being treated as errors


Another possible solution which doesn't work:



#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void test_function(char *a_intro, char *a_name);

int main(void)
char *my_intro;
char *my_name;
test_function(&my_intro, &my_name);

printf("String: %s %sn", my_intro, my_name);


void test_function(char *a_intro, char *a_name)

char *intro = malloc(20);
char *name = malloc(20);
strcpy(intro, "hello world");
strcpy(name, "James");

*a_intro = intro;
*a_name = name;




Error:



testcode.c: In function 'main':
testcode.c:10:19: error: passing argument 1 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c:10:30: error: passing argument 2 of 'test_function' from incompatible
pointer type [-Werror=incompatible-pointer-types]
test_function(&my_intro, &my_name);
^
testcode.c:5:6: note: expected 'char *' but argument is of type 'char **'
void test_function(char *a_intro, char *a_name);
^~~~~~~~~~~~~
testcode.c: In function 'test_function':
testcode.c:22:14: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_intro = intro;
^
testcode.c:23:13: error: assignment makes integer from pointer without a cast [-
Werror=int-conversion]
*a_name = name;
^
cc1.exe: all warnings being treated as errors




This question already has an answer here:



  • Update (int) variable in C inside a function [duplicate]

    5 answers



  • Change string with malloc in other function

    2 answers



  • Passing Char pointer in C function [duplicate]

    1 answer



  • How do I modify a pointer that has been passed into a function in C?

    5 answers







c string char malloc strcpy






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 15 '18 at 9:21









melpomene

62.4k55094




62.4k55094










asked Nov 15 '18 at 9:03









KiwaKiwa

416




416




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This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • What is unclear about the errors? test_function takes 2 arguments, you call it with 3. There is no variable my_str in main, you only have my_intro and my_name.

    – mch
    Nov 15 '18 at 9:06






  • 1





    In C all function arguments are passed by value. That means they are copied. The arguments inside a function are distinct and separate variables from what was used when calling the function. Modifying the copy inside the function will not modify the original variable. To solve that (and your error) please do some research about emulating pass by reference in C.

    – Some programmer dude
    Nov 15 '18 at 9:07











  • @mch its not. I tried the solution in that answer already. That answer is about ints. I tried what solution for my char. It didn't work

    – Kiwa
    Nov 15 '18 at 9:12












  • @Kiwa The fundamentals of pass-by-value don't change depending on the type of your variables.

    – melpomene
    Nov 15 '18 at 9:16











  • @melpomene I've edited my question with the logic of the duplicate.

    – Kiwa
    Nov 15 '18 at 9:17

















  • What is unclear about the errors? test_function takes 2 arguments, you call it with 3. There is no variable my_str in main, you only have my_intro and my_name.

    – mch
    Nov 15 '18 at 9:06






  • 1





    In C all function arguments are passed by value. That means they are copied. The arguments inside a function are distinct and separate variables from what was used when calling the function. Modifying the copy inside the function will not modify the original variable. To solve that (and your error) please do some research about emulating pass by reference in C.

    – Some programmer dude
    Nov 15 '18 at 9:07











  • @mch its not. I tried the solution in that answer already. That answer is about ints. I tried what solution for my char. It didn't work

    – Kiwa
    Nov 15 '18 at 9:12












  • @Kiwa The fundamentals of pass-by-value don't change depending on the type of your variables.

    – melpomene
    Nov 15 '18 at 9:16











  • @melpomene I've edited my question with the logic of the duplicate.

    – Kiwa
    Nov 15 '18 at 9:17
















What is unclear about the errors? test_function takes 2 arguments, you call it with 3. There is no variable my_str in main, you only have my_intro and my_name.

– mch
Nov 15 '18 at 9:06





What is unclear about the errors? test_function takes 2 arguments, you call it with 3. There is no variable my_str in main, you only have my_intro and my_name.

– mch
Nov 15 '18 at 9:06




1




1





In C all function arguments are passed by value. That means they are copied. The arguments inside a function are distinct and separate variables from what was used when calling the function. Modifying the copy inside the function will not modify the original variable. To solve that (and your error) please do some research about emulating pass by reference in C.

– Some programmer dude
Nov 15 '18 at 9:07





In C all function arguments are passed by value. That means they are copied. The arguments inside a function are distinct and separate variables from what was used when calling the function. Modifying the copy inside the function will not modify the original variable. To solve that (and your error) please do some research about emulating pass by reference in C.

– Some programmer dude
Nov 15 '18 at 9:07













@mch its not. I tried the solution in that answer already. That answer is about ints. I tried what solution for my char. It didn't work

– Kiwa
Nov 15 '18 at 9:12






@mch its not. I tried the solution in that answer already. That answer is about ints. I tried what solution for my char. It didn't work

– Kiwa
Nov 15 '18 at 9:12














@Kiwa The fundamentals of pass-by-value don't change depending on the type of your variables.

– melpomene
Nov 15 '18 at 9:16





@Kiwa The fundamentals of pass-by-value don't change depending on the type of your variables.

– melpomene
Nov 15 '18 at 9:16













@melpomene I've edited my question with the logic of the duplicate.

– Kiwa
Nov 15 '18 at 9:17





@melpomene I've edited my question with the logic of the duplicate.

– Kiwa
Nov 15 '18 at 9:17












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