在C中訪問動態分配的嵌套結構體

Question

結構如下，內部結構體位於外部結構體內部。這兩個結構都是動態分配的。當我嘗試通過引用來訪問內部結構來更改地址值時，會出現問題。

例如：

typedef struct{ 
    char address[32]; 
}inner; 

typedef struct{ 
    struct inner innerStruct; 
}outer; 

main(){ 

    int i = 0; 
    outer* outerArray; 
    outer* outerReference; 
    inner* innerReference; 

    /* create 20 outer structs */ 
    outerArray = malloc(20 * sizeof(outerArray*)); 

    /* for each outer struct, dynamically allocate 10 inner structs */ 

    for(i = 0; i < 10; i++) 
    { 
     outerReference = outerArray + i; 
     outerReference->innerStruct = malloc(10 * sizeof(outerReference->innerStruct); 
    } 

} 

如何訪問outerArray[3][innerStruct[4]，第四外結構的第五內部結構，並改變它的地址值？

Answer 1

如果您要使用malloc()，則聲明innerStruct爲inner *，而不是struct inner。

正如您聲明的那樣，創建outer時分配了inner的內存。

還請注意，由於您已使用typedef，因此在聲明該類型的變量時，不需要struct關鍵字。

這是你的代碼的修正版本，編譯和運行：

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

typedef struct { 
    char address[32]; // 32 chars are allocated when an inner is created 
} inner; 

typedef struct { 
    inner innerStruct; // innerStruct is allocated when an outer is created 
} outer; 

typedef struct { 
    inner *innerStruct; // innerStruct must be allocated explicitly 
} outer2; 

int main(int argc, char *argv[]) { 
    int i = 0; 
    outer *outerArray; 
    outer2 *outer2Array; 

    outer *outerReference; 
    outer2 *outer2Reference; 

    /* create 20 outer structs (should check for out-of-mem error) */ 
    outerArray = malloc(20 * sizeof(outer)); 

    for (i = 0; i < 10; ++i) { 
    outerReference = outerArray + i; // ptr to i'th outer 
    // Note: innerStruct.address bcz it's a structure 
    sprintf(outerReference->innerStruct.address, "outer struct %d", i); 
    } 

    /* create 20 outer2 structs */ 
    outer2Array = malloc(20 * sizeof(outer2)); 

    /* for each outer struct, dynamically allocate 10 inner structs */ 
    for (i = 0; i < 10; ++i) { 
    outer2Reference = outer2Array + i; 
    outer2Reference->innerStruct = malloc(sizeof(inner)); 
    // Note: innerStruct->address bcz it's a pointer 
    sprintf(outer2Reference->innerStruct->address, "outer2 struct %d", i); 
    } 

    /* print all the data and free malloc'ed memory */ 
    for (i = 0; i < 10; ++i) { 
    printf("outer: %-20s, outer2: %-20s\n", 
     outerArray[i].innerStruct.address, 
     outer2Array[i].innerStruct->address); 
     free(outer2Array[i].innerStruct); 
    } 
    free(outer2Array); 
    free(outerArray); 
    return 0; 
} 

Answer 2

打印第四外部結構的第五內strucuture成員address，將數據複製到它並打印一遍：

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

typedef struct{ 
    char address[32]; 
}inner; 

typedef struct{ 
    inner* innerStruct; 
}outer; 

int main() 
{ 
    int i = 0; 
    outer* outerArray; 
    outer* outerReference; 
    /* inner* innerReference; */ /* unused */ 

    /* create 20 outer structs */ 
    outerArray = malloc(20 * sizeof(*outerArray)); 

    /* for each outer struct, dynamically allocate 10 inner structs */ 

    for(i = 0; i < 10; i++) 
    { 
     outerReference = outerArray + i; 
     outerReference->innerStruct = malloc(10 * sizeof(*outerReference->innerStruct)); 
    } 

    printf("address: '%s'\n", outerArray[3].innerStruct[4].address); 

    strcpy(outerArray[3].innerStruct[4].address, "<emtpy>"); 

    printf("address: '%s'\n", outerArray[3].innerStruct[4].address); 

    return 0; 
} 

下次您發佈代碼時，請如此友好並進行編譯。