Why does copying a specific buffer size with memcpy and sprintf, prints more chars in new buffer tha-CodePudding

I have a general understanding question! Here is my code:

#include <stdio.h>
#include <unistd.h>
#include <stdlib.h>
#include <cstring>

int main() {
    // read user input
    char input[64] = {0};
    read(0, input, 64);
    printf("You've entered ");
    printf(input);

    char newbuf[128];
    char smallbuf[8];

    // copy into smallbuf 8 bytes of input
    memcpy(smallbuf, input, 8);

    // send smallbuf of 8 bytes as string into newbuf
    sprintf(newbuf, "%s", smallbuf);

    // print newbuf
    printf(&newbuf[0]);

    return 0;
}

The behavior I get with 7 chars is okay, it does print 7 chars:

$ gcc a.cpp -o a.out && ./a.out
1234567
You've entered 1234567
1234567
1234567

But with 8 chars it prints a lot more of them and I'm wondering why does it do this:

$ gcc a.cpp -o a.out && ./a.out 
12345678
You've entered 12345678
1234567812345678

Thank you for explaining me! :)

CodePudding user response：

Code is attempting to print a character array as if it was a string leading to undefined behavior. smallbuf[] does not certainly contain a null character, so it is not a string.
"%s" expects a matching pointer to a string.

Either account for a null character

char smallbuf[8 1];
memcpy(smallbuf, input, 8);
smallbuf[8] = '\0';
printf("%s", smallbuf);

or limit output with a precision. That prints a character array up to N characters or a null character.

char smallbuf[8];
memcpy(smallbuf, input, 8);
printf("%.8s", smallbuf);

Similar issue applies to printf(input);

Do not code printf(input); as that may lead to undefined behavior when input[] contains a %.

// printf(input);
printf("%s", input);

Better code would examine the return value of read(0, input, 64).