C++ 在同时使用 cout 和 printf 时没有四舍五入

Question

I need to make a program that calculates cos(x) and my problem is that when I use printf for example cos(0.2) is 0.98 but it comes out as 0.984 and it's not rounding to 2 numbers.

My code:

#include <iostream> 
#include <math.h>

using namespace std;

int main()
{
    float x = 0.2;
    cout << "x=" << x << " cos(y) y=" << printf("%.2f", cos(x)) << "\n";
    return 0;
}

Answer 1

The problem is not with rounding the numbers, but with output.

 cout << "x=" << x << " cos(y) y=" << printf("%.2f", cos(x)) << "\n";

Here you are mixing two ways to write to the standard output. Inserting a call to printf into cout << will output the return value of printf which happens to be 4 and at the same time output something as a side effect.

So two outputs are created:

• Streaming values into cout outputs x=0.2 cos(y) y=4
• Calling printf (correctly) outputs 0.98

It is possible that the two outputs are mingled with each other, creating the impression that the result was 0.984 :

x=0.2 cos(y) y=    4
               ^^^^
               0.98

You can use both cout and printf , but you should not confuse the return value of printf with the output it creates as a side effect :

cout << "x=" << x << " cos(y) y=";
printf("%.2f\n", cos(x));

should output

x=0.2 cos(y) y=0.98

See also: C++ mixing printf and cout

Answer 2

As others have said in the comments, mixing std::cout and printf does not do what you want. Instead use the stream manipulators std::fixed and std::setprecision :

#include <iomanip>  //Required for std::fixed and std::precision
#include <iostream> 
#include <cmath> //Notice corrected include, this is the C++ version of <math.h>

using namespace std;

int main()
{
    float x = 0.2f; //Initialize with a float instead of silently converting from a double to a float.
    cout << "x=" << x << " cos(y) y=" << std::fixed << std::setprecision(2) << cos(x) << "\n";
    return 0;
}