Question

我想在主要功能中实例化此“ JobSequence”结构。为此，我创建了一个new_JobSequence()函数，该函数分配内存并为结构的所有字段提供适当的值。在此函数中，一切都根据我的调试器正确完成。但是一旦回到主函数int cost就被设置为任意值baseInstance = 0x0和list = 0x1。 jobSequence之前和之后的new_JobSequence()地址是不变的，并且与函数内部使用的地址相同。这是结构的声明，函数的代码和main的代码：

typedef struct JobSequence{
    Instance* baseInstance;
    List* sequence;
    int cost;
}JobSequence;

void new_JobSequence(Instance* baseInstance, JobSequence* jobSequence){
    jobSequence = malloc(sizeof(JobSequence));
    jobSequence->baseInstance = baseInstance;
    jobSequence->cost = 0;
    list_new(&jobSequence->sequence);
}

int main() {
    int** P = baseInstance();
    Instance instance;
    makeInstanceFromBin(P,10,10,4,&instance);
    printInstance(&instance);
    JobSequence jobSequence;
    setbuf(stdout, 0);
    printf("address : %p\n",&jobSequence);
    new_JobSequence(&instance,&jobSequence);
    setbuf(stdout, 0);
    printf("address : %p\n",&jobSequence);

    Do other stuff (crashes because I try to access the elements of jobSequence which have 0x1 and 0x0 addresses).
}

所以我确实尝试了很多东西，但我真的不知道这里发生了什么...

谢谢您的帮助。

Answer 1

写作时

JobSequence jobSequence;

您要求编译器在堆栈上创建类型为JobSequence的变量。

所以您不必自己保留内存。

您在调试器中遇到的问题是jobSequence函数未修改main函数中的new_JobSequence。

您可以通过一些调试痕迹看到它：

void new_JobSequence(Instance* baseInstance, JobSequence* jobSequence){
    printf("start of %s: js is %p\n", __FUNCTION__, jobSequence);
    jobSequence = malloc(sizeof(JobSequence));

    printf("after malloc, js is %p\n", jobSequence);
    jobSequence->baseInstance = baseInstance;
    jobSequence->cost = 0;
    list_new(&jobSequence->sequence);       
}

因此，要使您的函数在main中创建的变量上起作用，您必须编写类似以下内容的

void new_JobSequence(Instance* baseInstance, JobSequence* jobSequence)
{
    jobSequence->baseInstance = baseInstance;
    jobSequence->cost = 0;
    list_new(&(jobSequence->sequence));
}

...
int main(void)
{
    /* ... */
    JobSequence jobSequence;
    new_JobSequence(..., &jobSequence);
}

如果要在堆上为结构分配内存，则必须遵循以下方法：

JobSequence *new_JobSequence(Instance* baseInstance){
    JobSequence *js= malloc(sizeof *js);
    js->baseInstance = baseInstance;
    js->cost = 0;
    list_new(&(js->sequence));

    return js;
}

...
int main(void)
{
    /* ... */
    JobSequence *jobSequence = new_JobSequence(...);                
}

Answer 2

您已经在主函数中分配了实例。因此，您实际上在new_JobSequence()内部所做的只是在堆中分配一些内存，对其进行初始化，并且在从该函数退出之前从不返回任何内容。如果要保持这种操作流程，则需要在函数中传递一个双指针（并在主函数（即JobSequence *pJobSequence = NULL中仅分配一个指针））。

C结构未初始化的外部构造函数

2 个答案: