Why extra value is printed using cout and printf?

Question

Why 5 is being printed after 1.00.

#include<iostream>
using namespace std;
int main() {
    double x = 1;
    cout << printf("%.2f\n", x);
    return 0;
}

output :

1.00

5

You could look up what [`printf`](https://en.cppreference.com/w/cpp/io/c/fprintf) does and, in particular, what it returns. — Pete Becker, May 01 '23 at 13:29
i know what printf does, but i cannot understand why 5 is getting outputted — ImBatman, May 01 '23 at 13:31
You could looke up what [`printf`](https://en.cppreference.com/w/cpp/io/c/fprintf) does and, in particular, **what it returns**. — sweenish, May 01 '23 at 13:32
Hint: [printf](https://en.cppreference.com/w/cpp/io/c/fprintf) returns the number of characters written. — wohlstad, May 01 '23 at 13:32
Can you use C++20? If so use [std::format](https://en.cppreference.com/w/cpp/utility/format/format) instead of printf (which is still kind of an "unsafe" leftover from C). format is faster and more typesafe then printf — Pepijn Kramer, May 01 '23 at 13:44
This question shouldn't have been closed as reproducible, voting to reopen. — vitaut, Jun 08 '23 at 13:53

Vlad from Moscow · Accepted Answer · 2023-05-01T13:40:45.740

From the C Standard (7.21.6.3 The printf function)

3 The printf function returns the number of characters transmitted, or a negative value if an output or encoding error occurred.

This call of printf

printf("%.2f\n", x)

outputs on console the following characters (shown as a string)

"1.00\n"

That is 5 characters including the new line character '\n'. The return value of the call is 5 according to the quote of the C Standard. And this value is outputted using the operator << for the standard output C++ stream std::cout.

Why extra value is printed using cout and printf?

1 Answers1