mem比较数组以获得匹配字节数

Question

int a[10];
int b[10];
memcmp(a, b, sizeof(int) * 10);

memcmp() only tells us which memory block is bigger/smaller since it just returns -1,0,+1. Is there a way to know number of matching elements in a[] and b[] just after which the mismatch occurs.

Ex: a = {1, 2, 3, 4, 5, 6}
    b = {1, 2, 4}

Here memcmp(a, b, sizeof(int)* 3) would return -1. I want to get 2 as answer Is there a way we can get number of matching elements with help of memcmp or some similar inbuilt function

Answer 1

I assume you want a low level answer since you must have already rejected the easy way with a loop.

You could XOR the memory block a[0] to a[9] with b[0] to b[9]. The resulting memory will be a block of zeros if, and only if, the original arrays were equal. To get the number of matching elements, count the number of int blocks that have all zero bits.

You could do this blisteringly fast in assembler as the operations are so simple.

Answer 2

A for loop over the two arrays should suffice:

int a[10];
int b[10];
int aLen = sizeof(a)/sizeof(int); /* we can do this since arrays are declared on the stack */
int bLen = sizeof(b)/sizeof(int);
int currentContig = 0;
int maxContig = 0;

for (int aIdx = 0, int bIdx = 0; (aIdx < aLen) && (bIdx < bLen); ) {
    if (a[aIdx] == b[bIdx]) {
        currentContig++;
        aIdx++;
        bIdx++;
    }
    else if (a[aIdx] > b[bIdx]) {
        currentContig = 0;
        aIdx++;
    }
    else {
        currentContig = 0;
        bIdx++;
    }
    if (currentContig > maxContig) {
        maxContig = currentContig;
    }
}

fprintf(stdout, "longest contiguous match: %d\n", maxContig);

Answer 3

Please Try Like below

int matched_num=0;
while(a[matched_num]==b[matched_num])
matched_num++;
printf("%d",matched_num);

final printed value will be total number of matched element in both arrays.