请问这个方案的工作？ [英] How does this program work?

问题描述

#include <stdio.h>

int main() {
    float a = 1234.5f;
    printf("%d\n", a);
    return 0;
}

它显示一个 0 ！这怎么可能？什么是道理？

It displays a 0!! How is that possible? What is the reasoning?

我特意放了％d个在的printf 语句学习的printf 。

I have deliberately put a %d in the printf statement to study the behaviour of printf.

推荐答案

这是因为％d个预计的 INT 但您提供一个float。

That's because %d expects an int but you've provided a float.

使用％E / ％F / ％G 打印浮动。

Use %e/%f/%g to print the float.

在为什么打印0：浮点数发送给的printf 之前转换成双击。在小尾数双重新presentation数1234.5为

On why 0 is printed: The floating point number is converted to double before sending to printf. The number 1234.5 in double representation in little endian is

00 00 00 00  00 4A 93 40

A ％d个消耗了32位整数，所以零被打印出来。 （作为一个测试，你可以的printf（％D，％d个\\ N，1234.5f）; 你可以得到输出 0，1083394560 ）

A %d consumes a 32-bit integer, so a zero is printed. (As a test, you could printf("%d, %d\n", 1234.5f); You could get on output 0, 1083394560.)

至于为什么浮动转换为双击，如printf的原型是 INT的printf（为const char *，...），从6.5.2.2/7，

As for why the float is converted to double, as the prototype of printf is int printf(const char*, ...), from 6.5.2.2/7,

在一个函数原型声明符引起的参数类型转换到最后声明参数后，停止省略号。 默认参数提升是在尾随参数执行。

The ellipsis notation in a function prototype declarator causes argument type conversion to stop after the last declared parameter. The default argument promotions are performed on trailing arguments.

和来自6.5.2.2/6，

and from 6.5.2.2/6,

如果它表示所调用的函数的前pression有一个类型不包括样机，整数促销活动每个参数执行，和参数具有类型浮动都提升到双击。这些被称为的默认参数提升的

If the expression that denotes the called function has a type that does not include a prototype, the integer promotions are performed on each argument, and arguments that have type float are promoted to double. These are called the default argument promotions.

（感谢阿洛克寻找这一点。）

(Thanks Alok for finding this out.)

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