为什么是负INT比unsigned int类型更大？ [英] Why is a negative int greater than unsigned int?

问题描述

  INT主要（无效）
{
     unsigned int类型Y = 10;
     INT X =  -  4;
     如果（X＆GT; Y）
        的printf（x大于）;
     其他
        的printf（Y大于）;
     残培（）;
     返回（0）;
}输出：x大于
 

我想输出是y是更大，因为它是无符号。是什么原因背后？

解决方案

由于该 INT 值提升到 unsigned int类型。特别是 0xFFFFFFFC 32位的机器，它作为一个 unsigned int类型是 4294967292 ，不是 10

C99 6.3.1.1-P2

  

如果int可以重新present（由宽度的限制，对于一个位字段）的原始类型的所有值，该值被转换为int;否则，将其转换为 unsigned int类型即可。这些被称为整数促销。所有其他类型是整数的提升不会改变。

要执行转换：

C99 6.3.1.3-P2

  

否则，如果新类型是无符号的值被重复地增加或减1比可以重新$ P $在新型psented直到该值是在新的类型的范围的最大值更转换

这基本上意味着增加UINT_MAX + 1（因为我读它，反正）。

至于为什么推广是对 unsigned int类型的一面; precedence：

C99 6.3.1.8-P1

  

...否则，如果具有无符号整型操作数的秩大于或等于另一个操作数的类型的秩，然后用符号整型操作数被转换为操作数的无符号整数的类型类型。

  
  

否则，如果有符号整数类型的操作数的类型可以重新present所有无符号整型操作数的类型的值，然后用无符号整型的操作数转换为的类型操作有符号整数类型。

这告诉我 INT 与 unsigned char型应正常工作。

测试

  INT的main（）
{
    INT X = -4;
    unsigned int类型Y = 10;
    unsigned char型Z = 10;    如果（X＆GT; Y）
        的printf（X＆GT; Y \\ n）;
    其他
        的printf（X＆LT; Y \\ n）;    如果（X将Z）
        的printf（X＆GT; Z \\ n）;
    其他
        的printf（X＆LT; Z \\ n）;
    返回0;
}
 

输出

  X'GT; Y
X'LT; Z
 

那么看看这个。

int main(void) 
{ 
     unsigned int y = 10; 
     int x = – 4; 
     if (x > y) 
        Printf("x is greater");
     else 
        Printf("y is greater"); 
     getch(); 
     return (0); 
} 

Output: x is greater

I thought the output would be y is greater since it is unsigned. What's the reason behind this?

解决方案

Because the int value is promoted to an unsigned int. specifically 0xFFFFFFFC on a 32-bit machine, which as an unsigned int is 4294967292, considerably larger than 10

C99 6.3.1.1-p2

If an int can represent all values of the original type (as restricted by the width, for a bit-field), the value is converted to an int; otherwise, it is converted to an unsigned int. These are called the integer promotions. All other types are unchanged by the integer promotions.

To perform the conversion:

C99 6.3.1.3-p2

Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.

Which basically means "add UINT_MAX+1" (as I read it, anyway).

Regarding why the promotion was to the unsigned int side; precedence:

C99 6.3.1.8-p1

...Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.

Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.

Which tells me int vs. unsigned char should work as expected.

Test

int main()
{
    int x = -4;
    unsigned int y = 10;
    unsigned char z = 10;

    if (x > y)
        printf("x>y\n");
    else
        printf("x<y\n");

    if (x > z)
        printf("x>z\n");
    else
        printf("x<z\n");
    return 0;
}

Output

x>y
x<z

Well look at that.

这篇关于为什么是负INT比unsigned int类型更大？的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！