Can not copy into char array from argument via strcpy

Question

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

void doSth(char *a)
{
    char *b,*c;
    b = malloc(2*sizeof(char));
    b[0]='a';
    b[1]='\0';

    a = malloc(2*sizeof(char));
    c = malloc(2*sizeof(char));
    strcpy(a,b);
    strcpy(c,b);

    printf("c:%s\n",c);
    free(c);
    free(b);
}


int main()
{
    char *myString;

    doSth(myString);

    printf("%s\n",myString);

    free(myString);
    return 0;
}

This program outputs only "c:a". Why can't I copy b to a? According to the debugger, the variable "a" remains empty in every line.

always check the returned value from malloc (and family) (!=NULL) to assure the operation was successful — user3629249
– user3629249, Commented Mar 21, 2015 at 23:15

francis · Accepted Answer · 2015-03-21 23:23:47Z

4

By doing a = malloc(2*sizeof(char)); in function void doSth(char *a), a local copy of a is modified in the scope of the function, but not out of it. If you wish to modify myString, a pointer to myString (&myString) should be given to the function. It is called passing by pointer :

char *myString;

doSth(&myString);

The prototype of the function is changed accordingly :

void doSth(char **a){
  ...
  *a = malloc(2*sizeof(char));
  strcpy(*a,b);
  ...
}

For the same reason, it is int a=1;printf("%d",a); (printf just needs the value of a), but int a;scanf("%d",&a); (scanf needs a pointer to a, to modify the value of a in the main).

edited Mar 21, 2015 at 23:23

answered Mar 21, 2015 at 23:16

francis

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