指向字符串指针的空指针

Question

compareStrings 将比较两个字符串。当 void 指针转换为 char 指针时，下面的代码可以正常工作。

int compareStrings(void *value1, void *value2) {
    const char *str1, *str2;

    str1 =  (char *)value1;
    str2 =  (char *)value2;

    return strcmp(str1, str2);
}

但是，当我将类型转换为指向 char 的指针时，代码会转储分段错误错误，我认为这是合理的，它不应该。

int compareStrings(void *value1, void *value2) {
    const char **str1, **str2;

    str1 =  (char **)value1;
    str2 =  (char **)value2;

    return strcmp(*str1, *str2);
}

谁能解释第二个函数的问题？

更新：

完整代码为：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>


int compare(void *value1, void *value2);
int binarySearch(void **array, int size, void *value,
             int(*compareFunc)(void *, void *)); 
int compareStrings(void *value1, void *value2);

int binarySearch(void **array, int size, void *value,
             int(*compareFunc)(void *, void *)) {
    int low, mid, high, compareResult;

    low = 0;
    high = size;
    while (low < high) {
        mid = low + ((high - low) / 2);
        compareResult = (*compareFunc)(array[mid], value);
        if (compareResult < 0) {
            low = mid + 1;
        } else {
            high = mid;
        }
    }

    return low;
}

int compareStrings(void *value1, void *value2) {
    const char *str1, *str2;

    str1 =  (char *)value1;
    str2 =  (char *)value2;     

    return strcmp(str1, str2);
}


int main() {
    int nElements, maxStringLen, index;
    char **stringArray;
    char *sToFind;

    nElements = 10;
    maxStringLen = 100;   
    sToFind = NULL;
    stringArray = malloc(sizeof(char *) * nElements);

    for (int i = 0; i < nElements; i++) {
        stringArray[i] = malloc(sizeof(char) * maxStringLen);
        sprintf(stringArray[i], "sample%d", i+1);
    }

    sToFind = "sample3";
    index = binarySearch((void **)stringArray, nElements, sToFind,         compareStrings);
    if (index >= nElements) {
        printf ("ERROR: value %s not found at index %d\n", sToFind, index);
    }else{
        printf("item found at index %d!\n", index);     
    }

    for(int i = 0; i < nElements; i++) {
        free(stringArray[i]);
    }
    free(stringArray);

    return 0;   
} 

Answer 1

当我将类型转换为指向 char 的指针时，代码会转储 我认为合理地不应该出现分段错误错误。

当您将指针转换为指针时，该操作不会影响转换后的指针。它仍然指向相同的内存。 打印输出3.

证明了这一点

1. Original string addresses:
Address of str1= 0x55bc799fea65
Address of str2= 0x55bc799fea6a

3. The addresses after casting `(char **)` do not change:
Address of  (char **)value1= 0x55bc799fea65
Address of  (char **)value2= 0x55bc799fea6a

铸造不是&或*操作！

当您通过* 进行取消引用时，该操作将取消引用原始指针的值。

结果不是你想要的：

4.
Address of *(char **)value1= 0x343332 00 34 33 32 31
Address of *(char **)value2= 0x31     00 35 34 33 32    

这些不是您期望的地址，而是字符串的值。

这些值不是有效地址！ 当您尝试将它们用作字符串地址时，您将在 strcmp 函数中遇到分段错误。

看看演示程序：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>

void print_addresses(char *nr, char *d1, void *a1, char *d2, void *a2)
{
    printf("%s.\nAddress of %s= %p\nAddress of %s= %p\n\n",nr, d1, a1, d2, a2);
}

int compareStrings1 (void *value1, void *value2)
{
    char *str1, *str2;

    str1 = (char *) value1;  
    str2 = (char *) value2;

    //2.
    print_addresses("2", "str1", str1, "str2", str2);

    return strcmp(str1, str2);
}

int compareStrings2(void *value1, void *value2) {
    char **str1, **str2;
    char **s1, **s2;    

    str1 =  (char **)value1;  //  warning: assignment from incompatible pointer type 
    str2 =  (char **)value2;  //  warning: assignment from incompatible pointer type 

   //3. Addresses after casting `(char **)` do not change:
    print_addresses( "3",  " (char **)value1", str1, " (char **)value2", str2);  // str1,str2 are still pointing to the original strings!
    //---------------------------------------------------------------------------------------------

    print_addresses( "4", "*(char **)value1", *str1, "*(char **)value2", *str2);  // '*' this dereferences the address to the value of the first character

    printf("(*str1)= %c\n", *str1);       //'*' this dereferences the address to the value of the first character
    printf("(*str2)= %c\n\n", *str2);       //'*' this dereferences the address to the value of the first character 

    // Now:

    s1 =  (char **) &value1;             // '&'' gives us pointer to pointer
    s2 =  (char **) &value2;             // '&'' gives us pointer to pointer

    //5.
    print_addresses( "5", " (char **) & value1" ,   s1, " (char **) & value2",   s2);    // pointer to pointer address 

    //6.
    print_addresses( "6", "*(char **) & value1",   *s1, "*(char **) & value2",  *s2);    // dereferencing pointer to pointer

    return strcmp(*s1, *s2);             // OK!
}

int main()
{
    char *str1 = "1234";
    char *str2 = "2345";  // 5 bytes

    //1.  Original string addresses:
    print_addresses("1", "str1", str1, "str2", str2);

    int res1 = compareStrings1(str1, str2);
    int res2 = compareStrings2(str1, str2);

   return 0; 
}

输出：

1.
Address of str1= 0x55bc799fea65
Address of str2= 0x55bc799fea6a

2.
Address of str1= 0x55bc799fea65
Address of str2= 0x55bc799fea6a

3.
Address of  (char **)value1= 0x55bc799fea65
Address of  (char **)value2= 0x55bc799fea6a

4.
Address of *(char **)value1= 0x3433320034333231
Address of *(char **)value2= 0x310035343332

(*str1)= 1
(*str2)= 2

5.
Address of  (char **) & value1= 0x7ffd061f56b8
Address of  (char **) & value2= 0x7ffd061f56b0

6.
Address of *(char **) & value1= 0x55bc799fea65
Address of *(char **) & value2= 0x55bc799fea6a

希望对你有帮助。

Answer 2

问题是您正在转换一个指针，而不是将其转换为指向指针的指针。

演员(char **) 只是改变了以下表达式的类型，它不转换它。所以原始指针和结果指针仍然是同一个指针（它们指向内存中的同一个地方），只是类型不同。所以两者都指向一个字符串。但是通过你的演员表你对编译器撒了谎：你告诉他你知道你在做什么，并且指针指向一个指向字符串的指针。

所以你得到一个分段错误：指针指向文本而不是指向文本的指针：你正在推迟一个字符串和哎呀。

你应该写：

str1 =  (char **)&value1;

注意：您的const char * 可能是错误的。您是说字符串是常量（并且只有第一个），而不是变量是常量（应该写成char * const）。