printf outputs too often

Question

#include <stdio.h>


void fillline(char *line, char c, int len){
    for(int i = 0; i<len-1; i++)
        line[i] = c;
    line[len-1] = '\n';
    line[len] = '\0';
}

int main() {
    
    char xs[4][30];
    fillline(xs[0], '-', 30);
    fillline(xs[1], '(', 30);
    fillline(xs[2], ')', 30);
    fillline(xs[3], 'Z', 30);

    printf("%s", xs[0]);
    printf("%s", xs[1]);
    printf("%s", xs[2]);
    printf("%s", xs[3]);
       
}

Hello, C-programming newbie here. I have a problem with the code above. I expect

-----------------------------
(((((((((((((((((((((((((((((
)))))))))))))))))))))))))))))
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

as command-line output generated from my code, but I get

-----------------------------
(((((((((((((((((((((((((((((
)))))))))))))))))))))))))))))
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
(((((((((((((((((((((((((((((
)))))))))))))))))))))))))))))
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
)))))))))))))))))))))))))))))
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
ZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

What am I doing wrong? I tried fflush(stdout) and setbuf(stdout, NULL) (although I am definitely not sure that this is a stdout-problem).

Thank you in advance.

You have not allowed room for a null-terminator to your strings. Consequently each one flows into the next. If you look carefully at the output pattern, first all 4 are printed, then 3, 2, and 1. — Weather Vane, Aug 02 '20 at 17:24
`line[len] = '\0'` should raise red flags. You can't assign to an index equal to the length of the string. `len - 1` is the last valid index. — Carcigenicate, Aug 02 '20 at 17:25

RobertS supports Monica Cellio · Accepted Answer · 2020-08-11T13:45:48.563

The problem is because of two reasons.

The strings ain't got a null terminator. Using it as string in printf("%s", xs[N]) causes undefined behavior.
With
```
line[len] = '\0';
```
you also attempt to write '\0' into an not existing 31th element.

Note that array indexing starts at 0, not 1. This confusion is what caused your trouble here.

Both things, the use of an indeterminate string and writing beyond the bounds of an array invokes undefined behavior.

Solution(s):

1.

Change

void fillline(char *line, char c, int len) {
    for(int i = 0; i < len-1; i++)
        line[i] = c;
    line[len-1] = '\n';
    line[len] = '\0';
}

to

void fillline(char *line, char c, int len) {
    for(int i = 0; i < len-2; i++)
        line[i] = c;
    line[len-2] = '\n';
    line[len-1] = '\0';
}

OR

2.

Decrement len before the use in the loop by 1.

void fillline(char *line, char c, int len) {
    len--;
    for(int i = 0; i < len-1; i++)
        line[i] = c;
    line[len-1] = '\n';
    line[len] = '\0';
}

OR

3.

Call the function fillline with an third argument of value 29, not 30.

fillline(xs[0], '-', 29);

score 3 · Answer 2 · answered Aug 02 '20 at 17:29

3

You are overwriting the end of your buffer.

void fillline(char *line, char c, int len){
    for(int i = 0; i<len-3; i++)
        line[i] = c;
    line[len-2] = '\n';
    line[len-1] = '\0';
}

answered Aug 02 '20 at 17:29

Paul Baxter

1,054
12
22

1

`< len - 3` needs to be `<= len - 3` OR `< len - 2`. Otherwise also UB. – RobertS supports Monica Cellio Aug 02 '20 at 17:59

score 1 · Answer 3 · edited Aug 02 '20 at 18:23

1

That is no '\0' at the end of string due to you are trying to write over the array, this will cause undefined behavior.

Sometimes it is a bad idea not performing bound checking in C.

edited Aug 02 '20 at 18:23

Roberto Caboni

7,252
10
25
39

answered Aug 02 '20 at 17:52

Tan Chun Hong

11
1

printf outputs too often

3 Answers3

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