C语言 printf 不打印到屏幕
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printf not printing to screen
提问by user1060986
If I try to run the following simple code under Cygwin on Windows 7,
如果我尝试在 Windows 7 上的 Cygwin 下运行以下简单代码，
#include <stdio.h>
int main() {
int i1, i2, sums;
printf( "Enter first integer\n" );
scanf( "%d", &i1 );
printf( "Enter second integer\n" );
scanf( "%d", &i2 );
sums = i1 + i2;
printf( "Sum is %d\n", sums );
return 0;
}
it compiles (via gcc) without a problem, but when I try to execute it, the first statement ("Enter first integer") isn't printed to the terminal, and I have to input two successive numbers (e.g. 3 and 4) before I get,
它编译（通过 gcc）没有问题，但是当我尝试执行它时，第一条语句（“输入第一个整数”）没有打印到终端，我必须输入两个连续的数字（例如 3 和 4）在我得到之前，
3
4
Enter first integer
Enter second integer
Sum is 7
Can anyone explain to me what is happening here. This works perfectly well under MinGW.
谁能向我解释这里发生了什么。这在 MinGW 下非常有效。
回答by zsawyer
Like @thejh said your stream seems to be buffered. Data is not yet written to the controlled sequence.
就像@thejh 说你的流似乎被缓冲了一样。数据尚未写入受控序列。
Instead of fiddling with the buffer setting you could call fflushafter each write to profit from the buffer and still enforce the desired behavior/display explicitly.
您可以fflush在每次写入后调用缓冲区设置，而不是摆弄缓冲区设置，以从缓冲区中获利，并且仍然明确地强制执行所需的行为/显示。
printf( "Enter first integer\n" );
fflush( stdout );
scanf( "%d", &i1 );
回答by Dayal rai
you can try for disabling the buffering in stdout by using
您可以尝试使用禁用标准输出中的缓冲
setbuf(stdout, NULL);
回答by thejh
It seems that the output of your program is buffered. Try enabling line buffering explicitly:
您的程序的输出似乎已被缓冲。尝试显式启用行缓冲：
setlinebuf(stdout);
