Question

初学者问题：我正在尝试创建一个包含函数的char数组，它应该比较2个数组 - 模式和文本，并返回最相似的文本的索引：

    int main(int argc, const char * argv[])
{

        int answer =0;

        answer  = match("viva","vviaaa");
        printf("%d",answer);
        return 0;


}

int match(char pattern[],char text[]) {
    int i,t,counter=0,topCount=0,topIndex=0;
    for (t=0;pattern[t]!= '\0'; t++) {
        counter=0;
        for (i=t; text[i]!='\0'; i++) {
            printf("is %c = %c ?\n",pattern[i],text[i]);
            if (pattern[i] == text[i]) {
                counter++;
                printf("yes\n");
            }
            if (pattern[i] =='\0') {
                break;
            }
        }
        if (counter>topCount) {
            printf("%d > %d at index t: %d\n",counter,topCount,t);
            topCount = counter;
            topIndex = t;

        }
    }
    return topIndex;
}

输出是（而不是blacnk char，有一个'？'倒置图片包括：

is v = v ?
yes
is i = v ?
is v = i ?
is a = a ?
yes
is � = a ?
2 > 0 at index t: 0
is i = v ?
is v = i ?
is a = a ?
yes
is � = a ?
is v = i ?
is a = a ?
yes
is � = a ?
is a = a ?
yes
is � = a ?
0Program ended with exit code: 0

Answer 1

我对您的代码进行了一些更改，现在可以正常运行了。你没有正确比较你需要的角色......

int match(char pattern[],char text[]);

int main(int argc, const char * argv[])
{

    int answer = 0;

    answer  = match("viva","vviaaa");
    printf("%d\n",answer);
    return 0;
}

int match(char pattern[],char text[])
{
    int i,j,k,counter=0,topCount=0,topIndex=0;
    for (i=0; text[i]!= '\0'; i++)
    {
        counter = 0;
        for (j=0,k=i; pattern[j]!='\0' && text[k]!='\0' && pattern[j] == text[k]; j++,k++,counter++);

        if(counter > topCount)
        {
            topCount = counter;
            topIndex = i;
        }
    }
    return topIndex;
}

比较2个char数组失败

1 个答案: