为什么编译器在带符号值分配给无符号整数时不会产生错误？ - C ++ [英] Why compiler is not giving error when signed value is assigned to unsigned integer? - C++

问题描述

我知道 unsigned int 不能保存负值。但下面的代码编译没有任何错误/警告。

  unsigned int a = -10; 
  

当我打印变量 时，如果无符号变量不能容纳带符号的值，为什么编译器允许他们编译而不给出任何错误/警告？

任何想法？

编辑

编译器：VC ++编译器

strong>解决方案

需要使用警告级别4。

解决方案

Microsoft Visual C ++：

警告C4245：'initializing'：
从'int'
int'，已签名/未签名的不匹配

在警告级别4上。

$ b b

G ++

提供警告：

-0x00000000a'到

$ h b $ b

gcc main.c -Wconversion

p>

警告：负整数隐式转换为无符号类型

请注意，-Wall不会启用此警告。

---

将你的警告水平提高。

I know unsigned int can't hold negative values. But the following code compiles without any errors/warnings.

unsigned int a = -10;

When I print the variable a, I get a wrong value printed. If unsigned variables can't hold signed values, why do compilers allow them to compile without giving any error/warning?

Any thoughts?

Edit

Compiler : VC++ compiler

Solution

Need to use the warning level 4.

解决方案

Microsoft Visual C++:

warning C4245: 'initializing' : conversion from 'int' to 'unsigned int', signed/unsigned mismatch

On warning level 4.

G++

Gives me the warning:

warning: converting of negative value -0x00000000a' to unsigned int'

Without any -W directives.

GCC

You must use:

gcc main.c -Wconversion

Which will give the warning:

warning: negative integer implicitly converted to unsigned type

Note that -Wall will not enable this warning.

---

Maybe you just need to turn your warning levels up.

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