用strcpy连接字符串

Question

Here is my code

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


int main() {

 char f[] = "First";
 char s[] = "Second";
 char *tmp = malloc(strlen(f) + strlen(s) + 2);
 strcpy(tmp, f);
 strcpy(tmp, s);
 printf("%s", tmp);
 free(tmp);
 return 0;
}

I'm trying to concatenate f and s. The problem is that tmp contains only "Second" as a array. What I miss here

Answer 1

strcpy将字符串复制到目标的开头，您需要strcat 。

Answer 2

The second strcpy overwrites the previous one. Both copy its content to the tmp pointer (at the start of it). You should use tmp+strlen(f) .

Or even better use strcat .

And even better use more secure methods like: strncpy , strncat , etc..

Answer 3

If you insist on using strcpy , your code should be slightly modified:

int main() {
    const char *f = "First";
    const char *s = "Second";
    char *tmp = malloc(strlen(f) + strlen(s) + 1);
    strcpy(tmp, f);
    strcpy(tmp+strlen(f), s);
    printf("%s", tmp);
    free(tmp);
    return 0;
}

You should consider using strncpy instead of strcpy for safety reasons. Also, strcat is a more conventional function for concatenating C string.

EDIT Here is an example of using strncpy instead of strcpy

#define MAX 1024

int main() {
    const char *f = "First";
    const char *s = "Second";
    size_t len_f = min(strlen(f), MAX);
    size_t len_s = min(strlen(s), MAX);
    size_t len_total = len_f + len_s;
    char *tmp = malloc(len_total + 1);
    strncpy(tmp, f, len_f);
    strncpy(tmp+len_f, s, len_s);
    tmp[len_total] = '\0';
    printf("%s", tmp);
    free(tmp);
    return 0;
}

Answer 4

You may want to use strcat instead of your second strcpy call, like this:

strcpy(tmp, f);
strcat(tmp, s);

Note also that allocating strlen(f) + strlen(s) + 1 bytes for tmp is sufficient no need to allocate strlen(f) + strlen(s) + 2 bytes. After concatenation, you'll get only one string, so only one null character is required.

Answer 5

using strcat() instead, which means append a string accroding to the MSDN doc.strcpy() just means copy a string. If you don't want to use strcat(), you should point out the position by using strncpy() or strcpy_s(). Please refer to the document.

Answer 6

The problem is that you copy the second string in place of the first one (the first parameter of strcpy() is where to copy the string) and this effectively overwrites the first string. Here's an idea of what you need:

size_t firstLen = strlen( f );
size_t secondLen = strlen( s );    
char *tmp = malloc(firstLen + secondLen + 1);
strcpy(tmp, f);
strcpy(tmp + firstLen, s);

This can be achieved by using strcat() , although that would lead to an extra scan along the copied string.

Answer 7

Here is the correct idiomatic safe way to do what you want:

size_t l = strlen(f);
char *tmp = malloc(l + strlen(s) + 1);
strcpy(tmp, f);
strcpy(tmp+l, s);

or:

size_t l = strlen(f) + strlen(s) + 1;
char *tmp = malloc(l);
snprintf(tmp, l, "%s%s", f, s);

I tend to prefer the latter unless you're writing embedded systems code where you want to avoid pulling in printf dependency.

Finally, note that you should be testing malloc for failure, and that it's useless and harmful to allocate memory and copy the strings if all you want to do is print them - you could just do the following:

printf("%s%s", f, s);