无法通过strcpy从参数复制到char数组中

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.

Answer 1

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).