C语言的printf输出中的附加“ 12”

Question

I wrote the following program to learn how \\0 character works in C programs:

#include <stdio.h>

int main(void) {
    char a[] = {'1','2','\0','2'};
    int i=0;
    for (i=0; i<4; i++){
        printf("%c\n",a[i]);
    }
    printf("%s",a);
    return 0;
}

Well, that's okay and I have the following output :

sh-4.3$ gcc -o main *.c                                                                                          
sh-4.3$ main                                                                                                     
1                                                                                                                
2                                                                                                                

2

But when I remove \\n character from printf command, I receive one additional 12 in output:

//. Same as above
    for (i=0; i<4; i++){
        printf("%c",a[i]);
    }
//. Same as above

Output is:

12sh-4.3$ gcc -o main *.c                                                                                        
sh-4.3$ main 
12212

While I think I must see this :

12sh-4.3$ gcc -o main *.c                                                                                        
sh-4.3$ main 
122

Note that I used this online compiler to compile above programs. It's GNU GCC v4.8.3

Answer 1

The extra 12 you are getting is from the printf("%s",a); line:

• Iterating through the loop prints out the characters 1 , 2 , null (nothing), and 2 .
• Printing out the string at a prints up until the null character ( \\0 ), so 12 .

That's why you get the output 12212 .

---

Printing with the newlines prints out 1 , 2 , new line, 2 , and 12 on separate lines.

In your example you are missing the final 12 because you don't have new line at the end so your shell prompt gets printed after it: notice in your question that you have 12sh-4.3$ at the beginning of one of the lines.