如何在功能之外释放 malloc

Question

无法解决此问题 - 我的编译器一直告诉我，我在使用free（指针）函数时遇到了一些麻烦。所以我不确定我的指针是否工作，但调试表明实际上一切正常。只有自由函数无法释放内存。

    #include <stdio.h>              //Bibliothek für input/output.
    #include <stdlib.h>             //For malloc
    #include <math.h>               //Bibliothek für matchematische Operationen.
    #include <iostream>             //Bibliothek für in/output in C++.
    #include <stdbool.h>            //Bibliothek für boolean
    //Prototypes
int* readNumbers(int size);
int sumUpNumbers(int* sumpointer, int size);
//Main function
int main()
{
    int arraySize;  //Size of the malloc-array
    int* pointer;   //pointer for storing of the malloc-address
    int total;      //variable for the sumUpNumbers function
    pointer = NULL; //point on zero
    //inform the user before getting a number from him
    std::cout << "Please give the size of array:" << std::endl;
    fflush(stdout); //free the output window
    //get a number for the size of array
    scanf("%d", &arraySize);
    //call the readNumbers function and store the first address of
    //the malloc-array in pointer
    pointer = readNumbers(arraySize);
    //call the sumUpNumbers function and store the number in total
    total = sumUpNumbers(pointer, arraySize);
    fflush(stdout); //free the output window
    //show the number from total
    printf("n total of the array:%d", total);
    //call the free function for making the memory of
    //the malloc-array free again
    free(pointer);
    fflush(stdin);  //free the keyboard buffer
    getchar();      //wait for a feedback from user
    return 0;       //return 0 to the machine in case if everything works well
}

//This function has a pointer extension because we want to work with the
//array outside of this function. We give the function a size of the array
//we want to build. The function builds an array and fills it with numbers
//and than gives us back the first address of the array.
int* readNumbers(int size)
{
    int* array;         //pointer for creating of malloc-array
    int i;              //counter
    //pointer for storing of the first address of the array
    int* helpPointer;
    array = NULL;       //set the pointers
    helpPointer = NULL; //              on zero
    //create the array
    array = (int *) malloc(size * sizeof(int));
    //check the value of the array to be sure that we have created
    //the array without errors
    if(array != NULL)
    {
        //store the first address of the malloc-pointer
        helpPointer = array;
        //give some value to all the parts of array
        for(i=0; i<=size; i++)
        {
            //inform the user
            printf("n give the %d. nummber of the array:n", i+1);
            fflush(stdout); //free the output window
            //read the value
            scanf("%d", array+i);
        }
        return helpPointer; //return the first address
    }
    //if something went wrong by creating of the array, do:
    else
    {
        //tell the user, what we computer does't have enough memory
        std::cout << "There is no place for saving the data in mamory";
        return 0;   //return with failure
    }
}
//The input of this function is a pointer with the address of the malloc-array
//from the readNumbers and the size of this array. The function adds all the numbers
//from the array and gives us the result of the additation back.
int sumUpNumbers(int* sumpointer, int size)
{
    int sum;    //variable for storing of total value
    int i;      //counter
    sum = 0;    //set the sum on zero before work with it
    //count all the values from the array
    for(i=0; i<=size; i++)
    {
        //count one number after another
        sum = sum + *(sumpointer+i);
    }
    return sum; //return the total value
}

Answer 1

for循环的限制是错误的。您正在写入数组末尾的一个位置，这可能会损坏内存，以便以后程序失败。将for循环更改为：

for(i=0; i<size; i++)

Answer 2

在readNumbers函数中，您有：

for(i=0; i<=size; i++)

但是数组只有 size 个元素长，所以只需将<=更改为 < ：

for(i=0; i < size; i++)

您在sumUpNumbers函数中遇到了同样的问题。但这很可能只会导致不正确的总和，尽管它在技术上是未定义的行为。

Answer 3

您的代码有几个问题：

1. fflush(stdin)是未定义行为的生成器。
2. 两个不正确的计数器：如果大小为 size ，则必须计算for(i = 0; i < size; i++)
3. 如果NULL array，您的int* readNumbers(int size)将返回int而不是int*。
4. 奇怪的C和C++混合，没有明显的理由使用cin和cout

除了写了三个明显的错误（1）和（2）和（3）之外，你还强迫自己使用C++编译器（4）来编译一些东西，其中99%是纯C代码。为什么？

如果您将cin和cout替换为适当的scanf()和printf()调用，则可以摆脱C++。因此，您可以使用 C 编译器。在这种情况下，请确保还修改malloc调用以符合C标准：

array = malloc(size * sizeof(int)); //no result casting!

然后你会得到100%的C代码，它更容易阅读，研究和调试。