C：铸造最小的32位整数（-2147483648）浮动给出正数（2147483648.0） [英] C: Casting minimum 32-bit integer (-2147483648) to float gives positive number (2147483648.0)

问题描述

我在做一个嵌入式项目时，我遇到了一些东西，我觉得很奇怪的行为。我设法重现它在codePAD（见下文）进行确认，但没有我的机器上的任何其他C编译器试试他们。

I was working on an embedded project when I ran into something which I thought was strange behaviour. I managed to reproduce it on codepad (see below) to confirm, but don't have any other C compilers on my machine to try it on them.

场景：我有一个的#define 为最负值的32位整数可以容纳，然后我尝试使用这个带有浮点值进行比较如下图所示：

Scenario: I have a #define for the most negative value a 32-bit integer can hold, and then I try to use this to compare with a floating point value as shown below:

#define INT32_MIN (-2147483648L)

void main()
{
    float myNumber = 0.0f;
    if(myNumber > INT32_MIN)
    {
        printf("Everything is OK");
    }
    else
    {
        printf("The universe is broken!!");
    }
}

codePAD链接： HTTP：//$c$cpad.org/cBneMZL5

要我来说，它看起来好像这本code应该工作正常，但让我吃惊的它打印出宇宙坏了！。

To me it looks as though this this code should work fine, but to my surprise it prints out The universe is broken!!.

这code隐式注塑 INT32_MIN 到浮动，但事实证明，这导致 2147483648.0 的浮点值（正！），即使浮点类型是完全有能力再presenting的 -2147483648.0 。

This code implicitly casts the INT32_MIN to a float, but it turns out that this results in a floating point value of 2147483648.0 (positive!), even though the floating point type is perfectly capable of representing -2147483648.0.

没有人有任何见解的这种行为的原因是什么？

Does anyone have any insights into the cause of this behaviour?

code SOLUTION ：由于史蒂夫·杰索普在他的答复中提到， limits.h中和 stdint.h 包含正确的（工作） INT 范围定义一切已经，所以现在我使用这些不是我自己的的#define

CODE SOLUTION: As Steve Jessop mentioned in his answer, limits.h and stdint.h contain correct (working) int range defines already, so I'm now using these instead of my own #define

问题/解决方案说明摘要：提供的答案和讨论，我觉得这是发生了什么事情的一个很好的总结（注：也看答案/评论，因为它们提供了更详细的说明）

PROBLEM/SOLUTION EXPLANATION SUMMARY: Given the answers and discussions, I think this is a good summary of what's going on (note: still read the answers/comments because they provide a more detailed explanation):

• 我使用的是C89编译器与32位长 S，所以比 LONG_MAX 较大的值，并且小于或等于 ULONG_MAX 其次→后缀的类型为无符号长


• （ - 2147483648L）实际上是一个一元 - 上的无符号长（见previous点）值： - （2147483648L）。这种否定操作'包装'的值周围是 2147483648 的无符号长值（因为32位无符号长活动的 0 的范围 - 4294967295 ）


• 这无符号长号的看起来的像预期的负 INT 价值时，它得到印刷作为一个 INT 或传递给函数，因为它是第一个得到强制转换为 INT ，它包裹了这一点-of范围 2147483648 各地 -2147483648 （因为32位 INT 活动的范围-2147483648到2147483647）


• 演员到浮动，但是，使用实际无符号长值 2147483648 转换，造成的浮点值 2147483648.0 。

• I'm using a C89 compiler with 32-bit longs, so any values greater than LONG_MAX and less or equal to ULONG_MAX followed by the L postfix have a type of unsigned long
• (-2147483648L) is actually a unary - on an unsigned long (see previous point) value: -(2147483648L). This negation operation 'wraps' the value around to be the unsigned long value of 2147483648 (because 32-bit unsigned longs have the range 0 - 4294967295).
• This unsigned long number looks like the expected negative int value when it gets printed as an int or passed to a function because it is first getting cast to an int, which is wrapping this out-of-range 2147483648 around to -2147483648 (because 32-bit ints have the range -2147483648 to 2147483647)
• The cast to float, however, is using the actual unsigned long value 2147483648 for conversion, resulting in the floating-point value of 2147483648.0.

推荐答案

在C89有32位长， 2147483648L 的类型 unsigned long int类型（见3.1.3.2整型常量）。所以一旦模运算已经应用到一元减运算， INT32_MIN 为正值2147483648类型无符号长。

In C89 with a 32 bit long, 2147483648L has type unsigned long int (see 3.1.3.2 Integer constants). So once modulo arithmetic has been applied to the unary minus operation, INT32_MIN is the positive value 2147483648 with type unsigned long.

在C99， 2147483648L 已键入长如果长大于32位，或长长否则（见6.4.4.1整型常量）。因此，有没有问题， INT32_MIN 为负值-2147483648型长或长长。

In C99, 2147483648L has type long if long is bigger than 32 bits, or long long otherwise (see 6.4.4.1 Integer constants). So there is no problem and INT32_MIN is the negative value -2147483648 with type long or long long.

与同样在C89 长大于32位， 2147483648L 已键入长和 INT32_MIN 是负的。

Similarly in C89 with long larger than 32 bits, 2147483648L has type long and INT32_MIN is negative.

我猜您使用的是C89编译器具有32位长。

I guess you're using a C89 compiler with a 32 bit long.

看它的方法之一是，C99修复了C89一个错误。在C99十进制字面没有 U 后缀的总是的签署型，而C89则可能根据其价值进行带符号。

One way to look at it is that C99 fixes a "mistake" in C89. In C99 a decimal literal with no U suffix always has signed type, whereas in C89 it may be signed or unsigned depending on its value.

你应该做的，顺便说一句，是包括 limits.h中，并使用 INT_MIN 的最小值一个 INT 和 LONG_MIN 为最低值长。他们有正确的值的和的所期望的类型（ INT_MIN 是 INT ， LONG_MIN 是长）。如果你需要一个精确的32位类型，然后（假设你的实现是2的补码）：

What you should probably do, btw, is include limits.h and use INT_MIN for the minimum value of an int, and LONG_MIN for the minimum value of a long. They have the correct value and the expected type (INT_MIN is an int, LONG_MIN is a long). If you need an exact 32 bit type then (assuming your implementation is 2's complement):

• 为code不具有可移植，你可以使用任何类型的您preFER这是正确的尺寸，并声称它是在安全方面。


• 为code具有可移植，搜索一个版本的C99头 stdint.h 上的C89编译器的工作原理，并使用 int32_t 和 INT32_MIN 从


• 如果一切都失败了，写 stdint.h 自己，并使用WiSaGaN的回答除权pression。它有键入 INT 如果 INT 至少为32位，否则长。

• for code that doesn't have to be portable, you could use whichever type you prefer that's the correct size, and assert it to be on the safe side.
• for code that has to be portable, search for a version of the C99 header stdint.h that works on your C89 compiler, and use int32_t and INT32_MIN from that.
• if all else fails, write stdint.h yourself, and use the expression in WiSaGaN's answer. It has type int if int is at least 32 bits, otherwise long.

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