未如预期一个简单的C程序的输出 [英] A simple C program output is not as expected

问题描述

我在做这个简单的程序和我的脑海里卡住了。

 ＃包括LT＆;＆stdio.h中GT;    诠释的main（）
    {
        int类型的签署;
        符号字符℃;
        unsigned char型B：
        B = 0xF4中;
        A =（签字）B： // 1）这个任务应该是相同的
        C =（签字）B：
        的printf（％d个\\ N，一）;
        一个= C; // 2）这一个
        的printf（％d个\\ N，一）;
        返回0;
    }一个= 244，A = -12
 

我试过，以确保任务是顺利，因为它可以，但我得到两个不同的答案，我希望无论是作为-12。什么小东西我丢失？

---

有关问题是

  {
    B = 0xF4中;
    C = 0xF4中;
    A =（符号的int）B： //试图挽救迹象，但得到244
    的printf（％d个\\ N，一）;
    一个= C; // SIGN保存在这里，并得到-3
    的printf（％d个，一）;
}
 

244，-3

所以我想从unsigned char类型保存迹象。这是可能以这种方式？

解决方案

 符号字符℃;
    unsigned char型B：
    B = 0xF4中;    / * ... * /    C =（签字）B：
 

乙的值是 0xF4中（ 244 ），但 C 键入（符号字符）只能容纳 -128 和 127 （在你的实现）。因此，当 244 分配给 C ，先把它转化为符号字符和C表示这个整数的转换是实现定义。 GCC 为大多数实现
只是环绕模 256 ，这就是为什么的ç值为 -12 。

下面是 GCC 文件：

http://gcc.gnu.org/onlinedocs/gcc/Integers- implementation.html

  

对于转换的类型的宽度N的值减小模2 ^ N至是内的类型的范围

I was doing this simple program and my mind got stuck

    #include<stdio.h>

    int main()
    {
        signed int a;
        signed char c;
        unsigned char b;
        b = 0xf4;
        a = (signed)b;       //    1)this assignment should be same 
        c = (signed)b;
        printf("%d\n",a);
        a = c;              //     2)as this one
        printf("%d \n",a);
        return 0; 
    }

a = 244, a = -12

I tried to make sure that assignment is as smooth as it can be but I am getting two different answers and I expected both as -12. What small thing I am missing?

---

The relevant question was

{
    b = 0xf4;
    c = 0xf4;
    a = (signed int)b;           //trying to save sign but get 244 
    printf("%d\n", a);
    a = c;                       //sign saved here and got -3 
    printf("%d", a);
}

244, -3

so i am trying to save the sign from an unsigned char. Is this possible in this manner?

解决方案

    signed char c;
    unsigned char b;
    b = 0xf4;

    /* ... */

    c = (signed)b;

The value of b is 0xf4 (244) but c type (signed char) can only hold values between -128 and 127 (in your implementation). So when 244 is assigned to c it is first converted to signed char and C says this integer conversion is implementation defined. gcc as most implementations just wraps around modulo 256 and that's why the value of c is -12.

Here is gcc documentation:

http://gcc.gnu.org/onlinedocs/gcc/Integers-implementation.html

"For conversion to a type of width N, the value is reduced modulo 2^N to be within range of the type"

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