碰到一个unsigned long的怪问题

[frequent]

unsigned char chTest[5]={0x54,0x5a,0xf7,0x14,0x00};
unsigned long nTime=chTest[0]<<24L+chTest[1]<<16L+chTest[2]<<8L+chTest[3];
上面计算出来的nTime竟然为0.

于是我又改成下面这样：
unsigned long nTemp=0;
unsigned char chTest[5]={0x54,0x5a,0xf7,0x14,0x00};
unsigned long nTime=(chTest[0]+nTemp)<<24L+(chTest[1]+nTemp)<<16L+(chTest[2]+nTemp)<<8L+(chTest[3]+nTemp);

算出来的nTime还是为0，请问这是为什么呢？

[frequent]

unsigned long n[4]={0x54,0x5a,0xf7,0x14}; unsigned long nTime=n[0]<<24+n[1]<<16+n[2]<<8+n[3]; 这样算出来的nTime竟然也为0.

[shenyi0106]

unsigned long nTime     = (xx[0] << 24) + (xx[1] << 16) + (xx[2] << 8) + xx[3];
printf("nTime=%ld\n", nTime);

这样写，结果就对了。 原因是+,-运算符的优先级高于<<,>>运算符

[worldy]

unsigned long nTime=(long)chTest[0]<<24L+(long)chTest[1]<<16L+(long)chTest[2]<<8L+(long)chTest[3]; 最简洁的做法是： unsigned char chTest[5]={0x54,0x5a,0xf7,0x14,0x00}; //0-3单元顺序颠倒一下 unsigned long nTime=*((long*)chTest);

[「已注销」]

强制转换一下：

BYTE chTest[] = { 0x54, 0x5A, 0xF7, 0x14 };
ULONG nTime = ((ULONG)chTest[0] << 24) | 
              ((ULONG)chTest[1] << 16) | 
              ((ULONG)chTest[2] << 8) | 
              ((ULONG)chTest[3]);

[赵4老师]

仅供参考：

#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

[Sandrer]

运算符优先级 你的运算符优先级没处理好 +-*/ 的优先级在 << >> 前，所以编译器会先计算 +, 再到 << {0x54,0x5a,0xf7,0x14,0x00} chTest[0]<<24L+chTest[1]<<16L+chTest[2]<<8L+chTest[3] = 84 << 114 << 263 << 28 这结果不溢出就算不错的了