Why the strcat function move the pointer to the next character?

Question

So, my problem is pretty simple, I don't know why the first segment of code does not work properly. The program read a string of 12 characters from a pipe and the strcat fuction moves the pointer of buff from the first character to the next every time the fuction is executed and so after a few interactions the read function make the program fail because the buffer is not big enough anymore. Using the sprintf function and another string solve the issue but i don't understand what cause the problem. Thanks for the help.

int n; 
char buff[15];
close(fd[1]);
    while(n = read(fd[0],buff,12) > 0){      
        strcat(buff,"\n");
        write(1,buff,13); 
        buff[0] = '\0'; 
        }

int n; 
char buff[15];
char output[15];
close(fd[1]);
while(n = read(fd[0],buff,12) > 0){      
            sprintf(output,"%s\n",buff); 
            write(1,output,13); 
            buff[0] = '\0';       
        }

`while(n = read(fd[0],buff,12) > 0)` is also wrong. Lookup the precedence of `=` and `>` — Paul Ogilvie, Apr 12 '21 at 14:02
And `write(1,output,13);`: why write 13 bytes if you first asked `n`? — Paul Ogilvie, Apr 12 '21 at 14:03
It is not the first character but that last character that is moved. — MikeCAT, Apr 12 '21 at 14:05
@PaulOgilvie `fd` is not declared here. How did you get the knowledge? — MikeCAT, Apr 12 '21 at 14:33
@PaulOgilvie `fd[1]` can be anything. `fd` is just some array the OP defined somewhere. It's possibly filled by [pipe](https://linux.die.net/man/2/pipe) — Kevin, Apr 12 '21 at 14:38

Paul Ogilvie · Accepted Answer · 2021-04-12T14:40:29.317

The proper code terminates the buffer, which is assumed to contain a string read:

int n;
char buff[15];
close(fd[1]);
while((n = read(fd[0],buff,12)) > 0){
    buff[n] = '\0'; /* add terminating null-character */
    strcat(buff,"\n");
    write(1,buff,n+1);
}

and

int n;
char buff[15];
char output[15];
close(fd[1]);
while((n = read(fd[0],buff,12)) > 0){
    buff[n] = '\0'; /* add terminating null-character */
    sprintf(output,"%s\n",buff);
    write(1,output,n+1);
}

Note the extra ( and ) in the assignment to n
Note the use of n, the actual number of characters read
And note, as Mike said, the termination of the string.

score -1 · Answer 2 · answered Apr 12 '21 at 14:03

-1

strcat and %s specifier of sprintf expects strings, which means "null-terminted sequence of characters" in C.

If the contents from the pipe is not guaranteed to null-terminated, you have to add terminating null-character.

int n;
char buff[15];
close(fd[1]);
while(n = read(fd[0],buff,12) > 0){
    buff[12] = '\0'; /* add terminating null-character */
    strcat(buff,"\n");
    write(1,buff,13);
    buff[0] = '\0';
}

int n;
char buff[15];
char output[15];
close(fd[1]);
while(n = read(fd[0],buff,12) > 0){
    buff[12] = '\0'; /* add terminating null-character */
    sprintf(output,"%s\n",buff);
    write(1,output,13);
    buff[0] = '\0';
}

answered Apr 12 '21 at 14:03

MikeCAT

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`buf[n]='\0';` you mean? – Paul Ogilvie Apr 12 '21 at 14:05
No need for `buff[0] = '\0'` I think (when using `n`) – Paul Ogilvie Apr 12 '21 at 14:06
@PaulOgilvie `buff[0] = '\0'` is not needed even without `n` (if `buff` is not used after the loop) because `buff[0]` will be overwritten by successful `read` anyway. – MikeCAT Apr 12 '21 at 14:08

Why the strcat function move the pointer to the next character?

2 Answers2