我有一个简单的C++程序,我认为会打印出f和g中定义的double值作为double ...但是C++将它们打印为整数.我检查了cout的默认精度,当我运行它时,输出显示默认精度应为6,所以我很困惑为什么双打不正确.
#include <iostream>
#include <stdio.h>
int main() {
double f = 735416.416898;
double g = f + 0.3;
std::cout << "default precision is " << std::cout.precision() << std::endl;
std::cout << "[c++] f = " << f << std::endl;
std::cout << "[c++] g = " << g << std::endl;
printf("[c] f = %f\n", f);
printf("[c] g = %f\n", g);
f = 735416.516898;
g = f + 0.3;
std::cout << "[c++] f = " << f << std::endl;
std::cout << "[c++] g = " << g << std::endl;
printf("[c] f = %f\n", f);
printf("[c] g = %f\n", g);
return 0;
}
我看到的输出是:
kiran@kiran-VirtualBox:~/test$ g++ -o test test.cpp
kiran@kiran-VirtualBox:~/test$ ./test
[c++] f = 735416
[c++] g = 735417
[c] f = 735416.416898
[c] g = 735416.716898
[c++] f = 735417
[c++] g = 735417
[c] f = 735416.516898
[c] g = 735416.816898
因为默认精度显示为6,cout是否也应该显示6位小数？
如果我做
std::cout << std::fixed << f
它打印正确(带有所有小数位).
只是想知道是否有人可以向我解释这种行为？这是预期的吗？我是否希望使用std :: fixed来强制打印双打？