关于产品scanf（）的使用LF符和printf（）与LF符？ [英] Question about scanf() with lf specifier and printf() with lf specifier?

问题描述

我学习C和我有以下的code：

I am learning C and I have the following code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

int main(int argc, char *argv[])
{
  double x;
  printf("x = ");
  scanf("%ld", &x);
  printf("x = %lf\n", x);

  system("PAUSE");  
  return 0;
}

（我用的开发C4.9中，Windows XP SP3）

(I am using Dev C4.9, Windows XP SP3)

当我运行上面的程序，进入5.3;该计划印刷 X = 0.000000

When I run the above program and entered 5.3; the program printed x = 0.000000

谁能解释这是为什么吗？

Can anyone explain why is that, please?

非常感谢。

推荐答案

的％LD 格式字符串意味着它预计在读长符号int ，但你传递它，而不是一个双击。您应该使用％LF 格式说明说，你想有一个双击。

The %ld format string means that it's expecting to read in a long signed int, but you're passing it instead a double. You should instead use the %lf format specifier to say that you want a double.

请注意，对于 scanf函数的→需要双击 S（并且要求是缺席浮动 S），而对于的printf 的→在％LF 没有效果：既％F 和％LF 有两个浮动 S中相同的输出和双击取值由于默认参数推广。

Note that for scanf, the l is required for doubles (and is required to be absent for floats), whereas for printf, the l in %lf has no effect: both %f and %lf have the same output for both floats and doubles, due to default argument promotion.

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