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Modifying a string in a function in C
I want to change a string in a function. My code is: void fillStr(char * string); int main() { ... char * str; ... fillStr(str); printf("%s", str); //the str doesn't contain what I want return 0; } void fillStr (char * string) { //A successful code with the malloc and the realloc callings for getcharing to the string. printf("%s", string); //outputs the string } Help me please to find out what I did wrong?
7 Answers
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Evgeniy Smelov, I've played around with this until it worked:
Seems you need double pointers.
This is my demonstration (input 'Hello'):
#include <stdio.h>
#include <stdlib.h>
void fillStr(char **string);
int main() {
char * str;
fillStr(&str);
printf("%s", str);
free(str);
return 0;
}
void fillStr (char ** string) {
*string = (char*)malloc(6);
for(int i=0; i<5; ++i)
(*string)[i] = getchar();
(*string)[5] = '\0';
printf("%s", *string);
}
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Martin, the fillStr's code successfully fills the string using the getchar() but it doesn't actually change the str. That's the question.
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The function works fine and it's printf() outputs exactly what I want. It gets a pointer on a string and works with it. Otherwise there would be a memory allocation error. The question is why the actual content of the main's str pointer doesn't change?
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Martin, (I don't know how to mention someone) yes but in the heap. So the HonFu's advice to use double pointer works fine. Thank you!