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C在非常基本的C程序上编程EXC_BAD_ACCESS

[英]C Programming EXC_BAD_ACCESS on very basic C program

I am quite new to C programming. 我是C编程的新手。 I need to write a programm that works with dynamic Arrays. 我需要编写一个与动态数组配合使用的程序。 (Takes in Values and doubles the size of the Array, when it is full). (当数据已满时,取值并使数组的大小加倍)。 My programm is finished and compilation works but I keep on getting this Error in the commented line: Thread 1: EXC_BAD_ACCESS (code=2, address:(some long adress). 我的程序已完成并且编译工作但我继续在注释行中得到此错误:线程1:EXC_BAD_ACCESS(代码= 2,地址:(一些长地址)。

I have read on this and it seems that I may be pointing to null. 我已经读过这个,似乎我可能指向null。 But a null test didn't do the job. 但是空测试没有完成这项工作。 I'Ve had this problem in other programms before and I seem to be missing a basic point. 我之前在其他程序中遇到过这个问题,我似乎错过了一个基本点。 Can somebody help me with this please! 请有人帮我这个! This is my code: 这是我的代码:

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

typedef struct{
    int *value;
    int size;
    int MAX;
} DynArray;

void dyn_array_add (DynArray* array){
    int wert;
    int *temp;
    printf("Geben Sie einen Wert ein:\n");
    scanf("%i", &wert);

    if (array->MAX==array->size) {
        for (int i= 0; i<array->MAX; i++) {
            temp[i] = array->value[i];  // error occurs HERE
        }
        free(array->value);

        array->MAX = array->MAX*2;
        array->value=malloc(sizeof(int)* array->MAX);
        for (int i= 0; i<array->MAX; i++) {
            array->value[i] = temp[i];
        }
    }

    array->value[array->size]= wert;
    array->size++;
    for (int i = 0; i < array->MAX; i++) {
        printf("Value[%i]: %i \n", i, array->value[i]);
    }
}

int main(int argc, const char * argv[]) {


    DynArray* array;
    array = (DynArray*)malloc(sizeof(DynArray));
    array->MAX=5;
    array->size=0;
    array->value=malloc(sizeof(int)* array->MAX);

    while (1) {
        dyn_array_add(array);

    }
    return 0;
}
temp[i] = array->value[i];  // error occurs HERE

显然,因为你没有为temp分配内存。

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