Question

这是我程序的一部分，当我运行此程序时，我遇到了分段错误。我把它缩小到这条线：

checkBase（ptr1，ptr2）

我将这两个作为指针传递。它们被声明为char *，它的运行时错误不是编译时间。文件包含

＆LT; a href =＆＃34; http://www.google.com＆＃34;＆gt; www.spam.google.com＆lt; / a＆gt;

在这种情况下，ptr1 = www.google.com和ptr2 = spam.google.com

while(inf){

    count++;
    getline(inf, line);
    //cout << "*******" << count << "*******" << endl <<line << endl;
    p = new char[line.length()+1];
    strcpy(p, line.c_str());
    if(strstr(p, "href")){

        ptr = strstr(p, "href");
        while(ptr[0]!='\0'){
            ptr += 1;
            if(ptr[0] == 'w' && ptr[1] == 'w' && ptr[2] == 'w'){
                cout << ptr << endl;            
                ptr = strtok(ptr, "\"");
                cout << "add1   " << ptr << endl;
                add1 = ptr;
                ptr1 = ptr;
                ptr = strtok(NULL, "> ");
                add2 = ptr;
                ptr2 = ptr;
                cout << "ptr1: " << ptr1 << endl << "ptr2: " <<ptr2 << endl;
                if(add1 == add2)
                    cout << "There is an exact match at line: " << count << endl << line << endl;
                else{
                    cout << "in else" << endl;
                    checkBase(ptr1, ptr2); //THIS GIVES A SEGMENTATION FAULT
                }
            }
        }
    }
}

void checkBase(char *add1, char *add2){
    cout << "here" << endl;
    char *base1[1000000], *base2[1000000];
    int count1 = 0, count2 = 0;
    base1[count1] = strtok(add1, ".");
    while(base1[count1] != NULL){
        count1++;
        base1[count1] = strtok(NULL, ".");
        cout << base1[count1] << endl;
    }
    base2[count2] = strtok(add2, ".");
    while(base2[count2] != NULL){
        count2++;
        base2[count2] = strtok(NULL, ".");
    }
    cout << base2[count2-1] << endl;
    if(((strcmp(base1[count1-1],base2[count2-1])) != 0) && (strcmp(base1[count1-2], base2[count2-2]) != 0)){
        //if((strcmp(base1[count1-1], base2[count2-1]) != 0)){
        cout << "Bases do not match: " << endl
             << base1[count1-2] << "." << base1[count1-1] << " and "
             << base2[count2-2] << "." << base2[count2-1] << endl;
        //}
    }
    else{
        cout << "Bases match: " << endl
             << base1[count1-2] << "." << base1[count1-1] << " and "
             << base2[count2-2] << "." << base2[count2-1] << endl;          
    }
}

我不知道为什么会出现分段错误。

Answer 1

char *base1[1000000], *base2[1000000];

毫无疑问，这会导致堆栈溢出。堆栈的大小有限，创建超过几kb的数组是一个坏主意。尝试在堆上分配它们，例如vector<char *> base1(1000000)

您还应该计算所需的确切大小并在向量上分配那么多或push_back。

Answer 2

除了@Neil Kirkwell已经提到的堆栈溢出之外的一些问题

那些不应该是while循环完全以base1 [count1]为条件！= NULL;你还应该确保count1小于数组中元素的数量。
如果count2或count1为0或1，您将尝试引用-1和-2的索引......不太好。
使用strrchr向后搜索，让您的生活更轻松
完全构建这些数组是浪费的，因为你似乎只关心最后两个标记，每个只需要两个指针。

即。

char *one_a = NULL, *one_b = NULL, *two_a=NULL, *two_b = NULL;
char *temp = strtok(add1, ".");
while (temp) {
    one_b = one_a;
    one_a = temp
    temp = strtok(NULL, ".");
}
char *temp = strtok(add2, ".");
while (temp) {
    two_b = two_a;
    two_a = temp
    temp = strtok(NULL, ".");
}
//now just compare one_a with two_a and one_b with two_b and you're done.

为什么这个程序在调用函数时会出现分段错误？

2 个答案: