如何在c中的结构上更改结构的元素？

Question

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

typedef char Name[100];

typedef struct Daughter
{
    Name name;
    int age;
    int height;
}Tochter;

typedef struct Father
{
    Tochter kleine[6];
    Name v;
    int age;
    int height;
    int numofdaugh;
}Dad;

int main (void)
{
    Vater Martin, *pDad;
    int numofdau;
    printf("How many Daughters? ");
    scanf("%i", Martin.numofdaugh);

    pDad = &Martin;
    pDad->kleine[0].name = "Alice"; 
    // I also tried Martin.kleine[0].name = "Alice";
}

I also tried to change elements with functions but it still didnt work and the error msg is: "expression must have pointer-to-object type but it has type "Tochter".

Answer 1

1. you cant assign arrays.
2. scanf requires pointer,.

    scanf("%i", &Martin.numofdaugh);

    pDad = &Martin;
    strcpy(pDad->kleine[0].name, "Alice"); 

PS Type Dad should be Vater

Answer 2

scanf needs to be passed a pointer.

Rather than:

scanf("%i", Martin.numofdaugh);

Pass a pointer to Martin.numofdaugh :

scanf("%i", &Martin.numofdaugh);

You also want to use strncpy to copy a string into a Name .

Name bobs_name = {0};
strncpy(bobs_name, "Bob", 100);

Using strncpy rather than strcpy prevents the possibility of an overflow, as a Name is limited to 100 characters.

Answer 3

pDad->kleine[0].name = "Alice";

is wrong because you can't copy C-strings this way (use strcpy ).

More correct is:

strcpy(pDad->kleine[0].name,"Alice"); 

Answer 4

@Chris 的答案是正确的，但是您还有一个其他地方不存在的 Type Vater 。