从 memcpy 打印结果的问题

Question

我正在尝试使用 memcpy 来挑选我感兴趣的特定数据范围，例如使用 Matlab array(100:200) 中的语法。然而，当我试图打印出结果来检查函数是否正确时，我得到了一个错误：Exception thrown at 0x00007FFF4A921399 (vcruntime140d.dll) in Examplefordebug.exe: 0xC0000005: Access violation writing location 0x0000000000000000. 有没有想过解决这个问题？结果假设是数组 ch2Buffer

中的两个 5

示例代码如下：

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

int main()
{
    int i,n;
    const int u32Size = 10;
    float* ch1Buffer = NULL;
    double* ch2Buffer = NULL;
    double* ch2newBuffer = NULL;
    
    int pBuffer[u32Size] = {10,2,10,2,10,5,10,5,10,2};
    int* pi16Buffer = pBuffer;

    ch1Buffer = (float*)calloc(u32Size, sizeof* ch1Buffer);
    ch2Buffer = (double*)calloc(u32Size, sizeof* ch2Buffer);
    

    // De-interveal the data to ch1Buffer and ch2Buffer
    for (i = 0; i < u32Size/2; i++)
    {
        ch1Buffer[i] += pi16Buffer[i * 2];
        ch2Buffer[i] += pi16Buffer[i * 2 + 1];
    }

    // Use memcpy to pick out the data we are interested
    memcpy(ch2newBuffer, &ch2Buffer[2], 2 * sizeof(ch2Buffer[0]));



    // Print out to check the result
    for (i = 0; i < 3; i++) {
        printf("%f ", ch2newBuffer[i]);
    }

    free(ch1Buffer);
    free(ch2Buffer);
    return 0;
}

Answer 1

"Access violation writing location 0x0000000000000000. Any thought to solve this problem?"

是的，您的代码正在尝试写入进程尚未拥有的内存位置...
你在这里有一个声明：

double* ch2newBuffer = NULL;

然后尝试将对象复制到它：

memcpy(ch2newBuffer, &ch2Buffer[2], 2 * sizeof(ch2Buffer[0]));

不先分配内存。分配内存将解决问题。

添加第三个内存分配，并在尝试使用缓冲区之前检查每个内存分配的结果。一些额外的建议，don't cast the return of (m)(c)(re)alloc() in C，并且在没有先验证调用是否成功之前不要使用用[m][c][re]alloc()创建的内存。这是包含以下建议的重构部分：

ch1Buffer = calloc(u32Size, sizeof* ch1Buffer);
if(ch1Buffer)
{
    ch2Buffer = calloc(u32Size, sizeof* ch2Buffer);
    if(ch2Buffer)
    {
        ch2newBuffer = calloc(u32Size, sizeof* ch2Buffer);
        if(ch2newBuffer)
        {
            // De-interveal the data to ch1Buffer and ch2Buffer
            for (i = 0; i < u32Size/2; i++)
            {
                ch1Buffer[i] += pi16Buffer[i * 2];
                ch2Buffer[i] += pi16Buffer[i * 2 + 1];
            }

            // Use memcpy to pick out the data we are interested
            memcpy(ch2newBuffer, &ch2Buffer[2], 2 * sizeof(ch2Buffer[0]));



            // Print out to check the result
            for (i = 0; i < 3; i++) {
                printf("%f ", ch2newBuffer[i]);
            }
            free(ch2newBuffer); 
        }else {/*handle error*/}
        free(ch2Buffer);
    }else {/*handle error*/}
    free(ch1Buffer);
}else {/*handle error*/}

一个额外的潜在问题...根据您的编译器，您可能在此处注册了一个额外的错误：

 const int u32Size = 10;
 ...
 int pBuffer[u32Size] = {10,2,10,2,10,5,10,5,10,2};

可变大小的对象VLA (defined in C99, and optionally in compilers since then.) may not be initialized at declaration. (read more about this here)

对于带有初始化器的非 VLA 数组，可以按以下方式进行声明，可以采用多种形式：

#define U32_SIZE 10

 int pBuffer[10] = {10,2,10,2,10,5,10,5,10,2};
 int pBuffer[U32_SIZE] = {10,2,10,2,10,5,10,5,10,2};
 int pBuffer[] = {10,2,10,2,10,5,10,5,10,2};

或者下面的VLA（无法初始化，只能在自动范围内声明。）

 int iBuffer[] = {10,2,10,2,10,5,10,5,10,2};//non-VLA for illustration
 int pBuffer[u32Size];//VLA 
 memcpy(pBuffer, iBuffer, sizeof(iBuffer);//update VLA with contents of iBuffer.

从 memcpy 打印结果的问题

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