C/C++语言 - POINTERS和ARITHMETIC ...例外代码:c0000005

Question

下面的代码编译正确,但是当它执行时,控制台显示以下错误...例外代码:c0000005.错误发生在以下行:

*cptr++ = hextbl[((tval >> 4) & 0x0F)];

此错误是关于不正确的内存访问.这样,我相信这个错误可能是我对指针和算术仍然不理解的东西......

#include <stdio.h>

// function prototypes
int main(int argc, const char *argv[]);
char *put_hexbyte(char *cptr, char tval);

// main routine
int main(int argc, const char *argv[]) // variables to get arguments
{
    char val = 65;     // 0x41 >>> I need 2 bytes 0x34 and 0x31,
                       // they are ASCII from 0x41 (0x34 = "4" and 0x31 = "1")
    char *bufASCII;    // pointer to store these ASCII

    bufASCII = put_hexbyte(bufASCII, val);

    return 0;
}


// Put a byte as hex ASCII, return pointer to next location.
char *put_hexbyte(char *cptr, char tval) 
{
    static char hextbl[16] =
    {
        '0', '1', '2', '3', '4', '5', '6', '7',
        '8', '9', 'A', 'B', 'C', 'D', 'E', 'F'
    };

    printf("at this point, all is OK!!!\n"); // <<< OK!
    *cptr++ = hextbl[((tval >> 4) & 0x0F)];  // <<< memory violation error! (Exception Code: c0000005)
    *cptr++ = hextbl[tval & 0x0F];

    return(cptr);
}

谢谢您的帮助!:)

Answer 1

你的指针:

char *bufASCII;

没有初始化.然后你写信给它:

*cptr++ = x;

您需要先将其初始化,否则使用它是未定义的行为.例如:

char *bufASCII = new char[2];

虽然那时候,这个:

bufASCII = put_hexbyte(bufASCII, val);

会失去原始指针的踪迹.如果你想要缓冲区末尾的返回值,你应该存储那个单独的:

char* eob = put_hexbyte(bufASCII, val);