could not pass a char array properly

Question

One function tries to pass char array to another, but without success:

char * ret() {
    char buf[32];
    buf[0]='1'; // buf char array contains 1
    buf[31]='\0';
    printf(buf);
    return buf;
}

int multiline() {
    char temp[32];
    strcpy(temp,ret()); // temp array doesn't contain 1 after this line
    temp[31]='\0'; 
    printf(temp);
}

Please tell me, how to fix this issue?

Why do you expect it to have 1? It's a bit unclear as to what you think is wrong with this code. Could you elaborate a bit more? — JaredPar, Feb 04 '12 at 18:46
I would like to copy buf array (which was returned by a ret() function) to the temp array. — Jake Badlands, Feb 04 '12 at 18:47
possible duplicate of [Pointer to local variable](http://stackoverflow.com/questions/4570366/pointer-to-local-variable) — Bo Persson, Feb 04 '12 at 19:23

score 2 · Answer 1 · answered Feb 04 '12 at 18:46

you are returning a pointer to a local variable char buf[32];. This array is allocated on the stack and only valid within the function but not outside of the function. You get undefined behavior when you access this array.

To return an array like this you should allocate it on the heap, eg. with malloc()

score 1 · Answer 2 · answered Feb 04 '12 at 18:48

You can't return a pointer to a local variable; as soon as the function returns, the local variable, along with the rest of the function's stack frame, is destroyed.

Instead, you can return a copy of the array; you could make one using strdup().

As an aside, you're setting the first character of your array, and putting a \0 at the end, but the 30 characters in between just hold garbage. Normally the \0 at the end of a string is immediately after the valid characters, not all the way at the end of the array.

sverre · Accepted Answer · 2012-02-04T18:57:13.393

1

You can fix the issue by not returning a local variable (allocate a string or return a pointer to a statically allocated one):

char * ret() {
    char * buf = malloc(32*sizeof(char));
    buf[0]='1'; // buf char array contains 1
    buf[31]='\0';
    printf(buf);
    return buf;
}

int multiline() {
    char temp[32];
    char * returnedString = ret();
    strcpy(temp,returnedString);
    free(returnedString);
    temp[31]='\0'; 
    printf(temp);
}

edited Feb 04 '12 at 18:57

answered Feb 04 '12 at 18:50

sverre

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The correct line is: `char * buf = (char *)malloc(32*sizeof(char));` but your answer really helped me! Thank you! – Jake Badlands Feb 04 '12 at 18:54
No, in C you do not usually cast the return value of malloc(), see http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc – sverre Feb 04 '12 at 18:55
With a line I mentioned earlier it compiles for me – Jake Badlands Feb 04 '12 at 18:58

could not pass a char array properly

3 Answers3