是否"cout<<(char *)NULL"进行"close(1)"操作这里? [英] Does "cout<<(char*)NULL" doing "close(1)" here?

问题描述

在以下代码中，在此行之后，我使用了 cout<<(char *)NULL; ，我的程序未在输出屏幕上打印任何内容.这是否意味着我在这里用 cout 完成了 close(1)?这里到底发生了什么?这是一个错误吗?请分享您的想法.

In the following code I have used cout<<(char*)NULL; after this line, my program printing nothing to the output screen. Does it mean I have done close(1) with cout here? What is actually happening here? Is this a bug? Please share your thoughts.

#include<iostream>
using namespace std;

void f(){
    cout<<"\nfun\n";
}

main(){
cout<<(char*)NULL;
f(); //not getting printed !
cout<<"\nhello\n";  //not getting printed !
cout<<"hii how are you?"; //not getting printed, why??
}

我已经使用gcc和DevCpp编译器尝试了此操作，观察到了相同的行为.

I have tried this with both gcc and DevCpp compilers, same behavior observed.

推荐答案

在这里，您在流上设置了 badbit ，这导致在 cout<<<(char *)之后不打印任何内容NULL;

Here you sets the badbit on the stream which causes nothing to be printed after cout<<(char*)NULL;

if (!__s)
 __out.setstate(ios_base::badbit);

该标准说: requires:不得为空指针.因此，您的程序肯定具有未定义的行为，应该对其进行修复.您可以使用 cout.clear()清除不良位.

The standard says: requires: shall not be a null pointer. So your program definitely has the undefined behavior and it should be fixed. You can clear the bad bit by using cout.clear().

在您的情况下， cout<<(char *)NULL; 会导致未定义的行为.但是GCC会安全播放.

In your case, cout<<(char*)NULL; causes undefined behavior. But GCC plays it safely.

希望这会有所帮助！

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