打印出串联字符串的长度

Question

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

void main()
{
  char a[8] = "hello";
  char b[8] = "HELLO";

  int x = strcmp(a,b);

   printf("%s \n%s \n", a, b); 

  printf("%s\n", strcat(a,b));
  printf("%d", strlen(strcat(a,b)));
}

This prints out:

hello

HELLO

helloHELLO

12

Why 12?

But if I have this:

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

 void main()
 {
  char a[8] = "hello";
  char b[8] = "HELLO";

  int x = strcmp(a,b);

  printf("%s \n%s \n", a, b); 

  //printf("%s\n", strcat(a,b));
  printf("%d", strlen(strcat(a,b)));
}

It prints out:

hello

HELLO

10

Could someone explain why the lengths are different here? I have a feeling that it's the '\\0' character though.

Answer 1

strcat modifies the string pointed to by the first argument. That is, the string represented by the second argument is appended to the string of the first argument.

Your char a[8] = "hello"; is 5 characters, plus a null, plus two unused bytes. Then b follows it in memory.

Before `strcat:

a:  h e l l o \0 _ _
b:  H E L L O \0 _ _

After first strcat :

a:  h  e  l  l  o  H  E  L
b:  L  O  \0 L  O  \0 _  _

Now a has overwritten into b , and b points to the string "LO" in essence.

Now the second strcat is concatenating "LO" to "helloHELLO". After second strcat :

a:  h  e  l  l  o  H  E  L
b:  L  O  L  O  \0 \0 _  _

Note the length of the string at a after the second strcat is 12 bytes: "helloHELLOLO".

In your second test case, you eliminated the first strcat , so you saw what looks like normal output. But, when you use strcat you must make sure your receiving buffer (the first argument) has enough room to hold the entire concatenated string plus null. When using C library functions it's advised to read the manual page. See man strcat for details.