printf的显示器很奇怪 [英] printf displays something weird

问题描述

有这样code：

#include <stdio.h>

int main() {
  float d = 1.0;
  int i = 2;
  printf("%d %d", d, i);
  getchar();
  return 0;
}

和输出是：

0 1072693248

我知道有printf和首次％d个错误应该以％F取代。但是，为什么变量i印错（的1072693248而不是2）？

I know that there is error in printf and first %d should be replaced with %f. But why variable i is printed wrong (1072693248 instead of 2)?

推荐答案

既然你指定的％d个而不是％F ，你真正看到的是 D 的二进制重新presentation为整数。

Since you specified %d instead of %f, what you're really seeing is the binary representation of d as an integer.

此外，由于数据类型不匹配，code实际上是未定义行为。

Also, since the datatypes don't match, the code actually has undefined behavior.

编辑：

现在来解释为什么你没有看到 2 ：

Now to explain why you don't see the 2:

浮动提升为双击在堆栈中。键入双击是（在本例中）8个字节长。然而，由于你的的printf 指定两个整数（在这种情况下，两个4字节），你看到的的二进制重新presentations 1.0 作为一种双击。 2.不打印，因为它超出了8个字节，你的的printf 预计。

float gets promoted to double on the stack. Type double is (in this case) 8 bytes long. However, since your printf specifies two integers (both 4 bytes in this case), you are seeing the binary representations of 1.0 as a type double. The 2 isn't printed because it is beyond the 8 bytes that your printf expects.

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