Question

我有以下代码。

#include <iostream>
using namespace std;
typedef unsigned char byte;

typedef  struct {
    unsigned int a;
    unsigned int b;
} Struct_A;

void* malloc_(size_t s, byte* heap, int *next) {
    int old = *next;
    *next = *next + s;
    return heap + old;
}

void f(Struct_A *sa, byte* heap, int *next) {
    sa = (Struct_A*) malloc_(8, heap, next);
    sa->a = 10;
    sa->b = 20;
}

int main() {
    byte *heap = new byte[1000];
    int next;
    next = 0;

    Struct_A *sa;
    sa = (Struct_A*) malloc_(8, heap, &next);
    sa->a = 100;
    sa->b = 200;

    cout << sa->a << endl;


    Struct_A *sb;
    f(sb, heap, &next);
    cout<< sb->a <<endl;


    return 0;

}

代码适用于sa但不适用于某人！函数f（）与＆＃34; Struct_A * sa之后的三个代码行完全相同;＆＃34;确实。知道函数f（）有什么问题吗？

Answer 1

你是某人，但你应该传递一个指向sb的指针并声明f以获得额外的间接：

void f(Struct_A **sa, byte* heap, int *next) {
    *sa = (Struct_A*) malloc_(8, heap, next);
    (*sa)->a = 10;
    (*sa)->b = 20;
}

int main() {

    Struct_A *sb;
    f(&sb, heap, &next);

在C ++

1 个答案: