Question

在更新用于计算开关表变量的柜台时,我陷入了奇怪的 b。

int iCount was assigned zero outside of the loop, and it is the counter used for the while loop.

为了更新通道内的反响,我写道:i。包装的海标+ = 包装的本案为7。然而,在浮标中,0+=包装结果是包装了Count+1, 即8。这导致整个坡道仍有一个阵列尚未填满。



当我把该线路改为icount=包装Count+iCount时,则将适当数值退还。

因此,这种行为是C所独有的,因为我经常在 Java这样做,没有奇怪的影响。

http://www.ohchr.org。 补充法律

#define SKIP   8
#define PACKED 7

int iCount;
iCount=0;

while (iCount < characters-1){  
    for (packedCount=iCount; packedCount< iCount+PACKED; packedCount++){ 
        //ASCII compressor logic goes here
    }
    //iCount+= packedCount; //this produces 8 for 0+packedCount

    //this works
    iCount= iCount+packedCount; //skip the next byte in array, since it was already packed
}

Answer 1

As far as the compilers are concerned

iCount += packedCount;
iCount = iCount + packedCount;

相同。如果它们重新产生不同的结果,那么你法典中的一些内容就会使iCount被拖走——也许是一个不好的参照。

Answer 2

ry , , ,

#define SKIP   8
#define PACKED 7

int iCount;
iCount=0;

while (iCount < characters-1){  
    for (packedCount=iCount; packedCount< iCount+PACKED; packedCount++){ 
        //ASCII compressor logic goes here
    }
    printf("iCount after the inner loop: %d
", iCount);             /* DEBUG */
    printf("packedCount after the inner loop: %d
", packedCount);   /* DEBUG */

    //iCount+= packedCount; //this produces 8 for 0+packedCount

    //this works
    iCount= iCount+packedCount; //skip the next byte in array, since it was already packed

    printf("iCount after update: %d
", iCount);                     /* DEBUG */
}

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