Question

当释放已分配的内存时程序崩溃：“在Y处修改的堆块在Y处理请求的大小为21”。

运行函数getUstr，在free（uStr）上崩溃。有人可以帮我找到我在哪里使用未分配的内存吗？谢谢！

int HexToUChars(char* hexStr, unsigned char **str){
    int i, n;
    int strLen = strlen(hexStr)/2;

    for (i = 0; i < (int)strLen; i++) {

        sscanf_s(hexStr + 2 * i*sizeof(unsigned char), "%02X", *str+i*sizeof(unsigned char));
    }

    return 0;
}

int getUStr(){
    char *hexStr = "E7CA7905DD060F0E437C885BF13DED9243B1D2BD94CB11223DA71360A8F7D2D4";
    unsigned char *uStr;
    size_t strLen = (size_t)strlen(hexStr) / 2;
    uStr = calloc((strLen + 1), sizeof(unsigned char));
    if (uStr != NULL){
        HexToUChars(hexStr, &uStr);
        free(uStr);//Error : Heap block at X modified at Y past requested size of 21
    }
}

Answer 1

对于sscanf()和朋友，%02X需要指向unsigned int的指针，但您指的是指向unsigned char的指针。如果您的系统上有一个unsigned int，例如4个字节，那么您将通过该循环在最后几次迭代中写入已分配内存的末尾。

您应该提供指向（本地）unsigned int到sscanf_s()的指针，然后将其值分配给您的unsigned char。

Answer 2

there were several errors in the presented code.
those errors are fixed here
I did add a few $includes, etc so the file would compile


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

int HexToUChars( const char *, unsigned char * );
int getUStr( void );

int HexToUChars(const char* pHexStr, unsigned char *pStr)
{
    int i = 0; // loop index
    //int n;  // this raises a compiler warning about unused variable
    // the strLen is using a reserved word, with only a capitalization change,
    // I.E. very poor program practice
    int hexStrLen = (int)strlen(pHexStr) >> 1;  // this assumes that hexStr is an even number of bytes

    for (i = 0; i < hexStrLen; i++)
    {
        pStr[i] =  pHexStr[2*i] - '0';
        pStr[i] += pHexStr[(2*i)+1] - '0';
    } // end for

    return 1;
} // end function: HexToUChars

int getUStr()
{
    const char* pHexStr = "E7CA7905DD060F0E437C885BF13DED9243B1D2BD94CB11223DA71360A8F7D2D4";
    unsigned char *pStr;

    int  hexStrLen = strlen(pHexStr) / 2;

    if( NULL == (pStr = calloc( (hexStrLen + 1), sizeof(unsigned char) ) ) )
    {
        perror( "calloc failed" );
        exit( EXIT_FAILURE );
    }

    // implied else, calloc successful


    HexToUChars(pHexStr, pStr);
    if( strlen((const char*)pStr) != hexStrLen )
    {
        perror( "problem in HexToUChars");
    }
    printf( "%s", pStr );

    free(pStr);

    return(0); // without this, compiler raises warning about no return statement
}

X处的堆块在Y处修改过去请求的大小 - 释放分配的内存

2 个答案: