在第一個函數中傳遞指向多個函數和內存分配的指針

Question

我的問題是在C編程和NOT C++的環境中！ 我想通過多個函數之間的指針。但是內存分配不應該由調用者完成。我嘗試了一個小例子來模擬這個。可以看出，當指針指向一個在main函數中定義的結構變量時，它就像預期的那樣工作。這是我的功能可以操縱該地址的值通過時，在該內存地址中的值。但是，當函數調用返回並且控制傳遞給main時，爲什麼指針會「重新初始化」？指針可以反映它指向的地址嗎？

這怎麼辦？

以下是我有：

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

//example to pass a struct to void pointer and a void pointer to struct 
//testing if memory allocation is done by callee and not caller 

typedef struct mystructure{ 
    int a; 
    char b; 
    unsigned char c[10]; 
}mystruct; 


void func(void *var){ 

    mystruct *s = var; 
    s->a = 100; 
    s->b = 'I'; 
    strncpy(s->c,"test",5); 
    printf("In func\n"); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 
} 

void voidOut(void *var){ 

    mystruct *s = var; 
    printf("In voidOut\n"); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 

} 

//here is void pointer is both and 'in' and 'out' parameter 
void memfunc(void *var){ 

    mystruct *s = var; 
    s = (mystruct *)malloc(sizeof(mystruct)); 
    s->a = 100; 
    s->b = 'I'; 
    printf("In memfunc\n"); 
    strncpy(s->c,"test",5); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 
} 

//here is void pointer is an 'in' parameter 
void memvoidOut(void *var){ 

    mystruct *s = var; 
    printf("In memvoidOut\n"); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 
} 


int main(int argc, char *argv[]){ 
    mystruct val; 
    func(&val); 
    voidOut(&val); 

    mystruct *ptr = NULL; 
    memfunc(ptr); 
    memvoidOut(ptr); 

    return 0; 
} 

UPDATE： 繼答案和評論，這裏就是我：

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

//example to pass a struct to void pointer and a void pointer to struct 
//testing if allocation is done by callee and not caller 

typedef struct mystructure{ 
    int a; 
    char b; 
    unsigned char c[10]; 
}mystruct; 


void func(void *var){ 

    mystruct *s = var; 
    s->a = 100; 
    s->b = 'I'; 
    strncpy(s->c,"test",5); 
    printf("In func\n"); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 
} 

void voidOut(void *var){ 

    mystruct *s = var; 
    printf("In voidOut\n"); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 

} 


//here is void pointer is both and 'in' and 'out' parameter 
void memfunc(void **var){ 

    mystruct *s = var; 
    s = (mystruct *)malloc(sizeof(mystruct)); 
    s->a = 100; 
    s->b = 'I'; 
    printf("In memfunc\n"); 
    strncpy(s->c,"test",5); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 
    //memcpy(var,s, sizeof(s)); 
} 

//here is void pointer is an 'in' parameter 
void memvoidOut(void **var){ 

    mystruct *s = var; 
    printf("In memvoidOut\n"); 
    printf("s->a = %d\n",s->a); 
    printf("s->b = %c\n",s->b); 
    printf("s->c = %s\n",s->c); 
} 


int main(int argc, char *argv[]){ 
    mystruct val; 
    func(&val); 
    voidOut(&val); 

    mystruct *ptr = NULL; 
    memfunc(&ptr); 
    memvoidOut(&ptr); 

    return 0; 
} 

但是我的輸出是：

In func 
s->a = 100 
s->b = I 
s->c = test 
In voidOut 
s->a = 100 
s->b = I 
s->c = test 
In memfunc 
s->a = 100 
s->b = I 
s->c = test 
In memvoidOut 
s->a = 0 
s->b = d 
s->c = 

我錯過了什麼？ Shoud爲memvoidOut中的結構定義內存？

Answer 1

你應該改變這種分配新的存儲功能的簽名：

void memfunc(void *var) 

到

void memfunc(void **var) 

，並從主

memfunc(&ptr) 

解釋稱：

您的memfunc函數會複製傳遞的指針ptr。最初，它將此指針指定給分配的空間（通過malloc），但在函數返回後，指針仍然是傳遞給memfunc的原始指針（即NULL），因爲ptr指針的副本已被回收。

現在，如果你改變它的提議，該指針的地址將被傳遞，因此你會不會面臨「複製指針」的問題

Answer 2

函數參數通過值C傳遞。因此，您不能傳遞指針本身並從被調用函數分配內存到指針本身。

您可以執行兩件事情

1. 傳遞指針的地址從另一個函數分配的內存。

2. 通過指針，分配內存，從函數返回分配的指針並在調用者中收集返回值的同一指針。

所以，在memfunc()情況下，無論是

• 您必須更改簽名傳遞&ptr，在mystruct **var收集，然後在函數內部，分配內存*var

• 您需要退回mystruct *並收集ptr。