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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