c语言中右移问题

swordtan 2012-09-27 05:19:58



/*

 * main.c

 *

 *  Created on: 2012-9-27

 */



#include <stdio.h>



int main(){

	int a = 0x80000000, b, c;

	b = (a & 0xFFFFFFFF) >> 1;

	printf("%d\n", b);

	c = (a | 0x0) >> 1;

	printf("%d\n", c);

	return 0;

}

输出结果：
1073741824
-1073741824

为什么 (a & 0xFFFFFFFF) >> 1 是逻辑右移
而(a | 0x0) >> 1; 是算数右移呢？

...全文

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赵4老师 2012-09-28

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#include <stdio.h>

unsigned short int ui;

  signed short int si;

int main() {

    ui=(unsigned short int)0x8000u;

    si=(  signed short int)0x8000;

    printf("ui=%u\n",ui);

    printf("si=%d\n",si);

    ui=ui>>1;

    si=si>>1;

    printf("ui=%u\n",ui);

    printf("si=%d\n",si);



    printf("--------------\n");

    ui=(unsigned short int)0x8000u;

    si=(  signed short int)0x8000;

    printf("ui=%u\n",ui);

    printf("si=%d\n",si);

    ui=((  signed short int)ui)>>1;

    si=((unsigned short int)si)>>1;

    printf("ui=%u\n",ui);

    printf("si=%d\n",si);



    return 0;

}

//ui=32768

//si=-32768

//ui=16384

//si=-16384

//--------------

//ui=32768

//si=-32768

//ui=49152

//si=16384

mymtom 2012-09-28

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[Quote=引用 7 楼的回复:]
这是哪里的说明？
[/Quote]
C89
[url=http://www.bsb.me.uk/ansi-c/ansi-c-one-file.html][url]

ouPuso 2012-09-28

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0x00UL

AnYidan 2012-09-28

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[Quote=引用 6 楼的回复:]

是0xFFFFFFFF的类型超过了有符号的范围，使得数据类型提升为无符号的结果
lz可以这样测试一下:
printf("%#x, %#x\n",(a&(-1))>>1, (a&0xffffffff)>>1);
[/Quote]

If an int can represent all the values of the original type, then the value is converted to int; otherwise the value is converted to unsigned int. This process is called integral promotion.

swordtan 2012-09-27

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[Quote=引用 6 楼的回复:]

是0xFFFFFFFF的类型超过了有符号的范围，使得数据类型提升为无符号的结果
lz可以这样测试一下:
printf("%#x, %#x\n",(a&(-1))>>1, (a&0xffffffff)>>1);
[/Quote]

你的意思是，文字整型常量是有类型的，如：
0x0 ~ 0x7FFFFFFF int类型
0x80000000 ~ 0xFFFFFFFF 是unsigned int类型的,
对吧？

AndyZhang 2012-09-27

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[Quote=引用 3 楼的回复:]
The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient……
[/Quote]
这是哪里的说明？

ouPuso 2012-09-27

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是0xFFFFFFFF的类型超过了有符号的范围，使得数据类型提升为无符号的结果
lz可以这样测试一下:

printf("%#x, %#x\n",(a&(-1))>>1, (a&0xffffffff)>>1);

mujiok2003 2012-09-27

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hu7324829 2012-09-27

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算术位移只是针对有符号数有不同.



c = (a | 0x0u) >> 1;

如果提升a成无符号类型, 将会进行逻辑右移

mymtom 2012-09-27

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The result of E1 >> E2 is E1 right-shifted E2 bit positions. If E1 has an unsigned type or if E1 has a signed type and a nonnegative value, the value of the result is the integral part of the quotient of E1 divided by the quantity, 2 raised to the power E2. [B]If E1 has a signed type and a negative value, the resulting value is implementation-defined[/B].

图灵狗 2012-09-27

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参与位运算时，最好用无符号类型:



/*

 * main.c

 *

 *  Created on: 2012-9-27

 */



#include <stdio.h>



int main(){

    unsigned int a = 0x80000000, b, c;

    b = (a & 0xFFFFFFFF) >> 1;

    printf("%d\n", b);

    c = (a | 0x0) >> 1;

    printf("%d\n", c);

    return 0;

}