Why "$" is printed when I try to run the C code below?

Question

#include <stdio.h>
int main()
{
printf("%c\n","\"\"\"\"\"");

}

This prints $ but not """""

Why "$" is printed when I try to run the C code above?

`"%c"` is for a *single* `char`. Mismatching format specifier and argument leads to [*undefined behavior*](https://en.wikipedia.org/wiki/Undefined_behavior). UB (Undefined Behavior) means there's no use in discussing behavior. — Some programmer dude, Jul 14 '18 at 14:33
[What is the difference between %c and %s in C](https://stackoverflow.com/questions/33856654/what-is-the-difference-between-c-and-s-in-c/33856673) — user202729, Jul 14 '18 at 14:33

score 1 · Answer 1 · answered Jul 14 '18 at 14:44

1

You pass a string to printf, not a char. Try '"'

What happens in your code is that the string is passed to printf, but printf expects a char. printf now tries to interpret the address of this string as a char. In general, this is undefined behavior.

answered Jul 14 '18 at 14:44

Paul Ogilvie

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score 0 · Answer 2 · answered Jul 14 '18 at 14:34

0

%c is the format specifier for a single character; there is probably some conversion going on. You want %s (for strings).

answered Jul 14 '18 at 14:34

Anubhav Srivastava

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score 0 · Answer 3 · answered Jul 14 '18 at 14:36

0

It actually prints some garbage, because your input is not a single character, is a sequence of characters. Try using %s instead of %c

answered Jul 14 '18 at 14:36

Bogdan Cosma

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4

Why "$" is printed when I try to run the C code below?

3 Answers3