copying part of the char array to a string in c

Question

I have read many suggested questions, but still cannot find out the answer. I know the content in buffer is a NULL terminated char array, and I want to copy it into a dynamic allocated char array. However, I kept getting segmentation fault from the strcpy function. Thanks for any help.

void myFunction()
{
    char buffer[200];

    // buffer was filled by recvfrom correctly, and can be printed out with printf()
    char *message = malloc(200);

    strcpy(message, buffer[1]);
}

////////////////

ok, so i tried strcpy(message, &buffer[1]); strcpy(message, buffer); but nothing worked!!

crank up the warning level of your compiler and mind the warnings! — pmg
– pmg, Commented May 31, 2011 at 22:51
The second argument of strcpy() should also be a [const] char *, the buffer. It is currently the second item of the buffer, a char. — Jeff Mercado
– Jeff Mercado, Commented May 31, 2011 at 22:51
It is more idiomatic to write "strcpy( message, buffer + 1 )" — William Pursell
– William Pursell, Commented Jun 1, 2011 at 4:53
Your segfault is being caused by something different now. Not all segfaults are created equal. I updated the answer to include a terminating null byte in your buffer (the most likely candidate for what's causing your program to crash). Paste a larger snippet if you want more help or suggestions. — Sean
– Sean, Commented Jun 1, 2011 at 20:37

tofutim · Accepted Answer · 2011-05-31 23:17:50Z

2

This works for me. Is it possible that your buffer is not null-terminated?

char buffer[200];
buffer[0] = 'h';
buffer[1] = 'e';
buffer[2] = 'l';
buffer[3] = 'l';
buffer[4] = 'o';
buffer[5] = '\0';

// buffer was filled by recvfrom correctly, and can be printed out with printf()
char *message = (char *)malloc(200);
strcpy(message, buffer);

edited May 31, 2011 at 23:17

answered May 31, 2011 at 22:56

tofutim

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1 Comment

derrdji Over a year ago

I could do printf("message is: %s", buffer); and it correctly printed out the message I sent using sendto. I am sure buffer contains the right c string

Sean · Accepted Answer · 2011-06-01 20:36:24Z

1

Your invocation of strcpy(3) is incorrect. Change it to the following:

    buffer[199] = '\0';
    strcpy(message, &buffer[1]);

strcpy(3) has the following signature:

 char *
 stpcpy(char *s1, const char *s2);

You passed in:

 char *stpcpy(char *s1, const char s2); /* won't work */

I would suggest using memcpy(3) instead of strcpy(3) since strcpy(3) relies on a null character to terminate the string.

edited Jun 1, 2011 at 20:36

answered May 31, 2011 at 22:55

Sean

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