Initialising an array of structures in C

Question

I'm doing a simple student database program exercise and I'm unsure of how to initialise an array of structures. I'm trying to initialize the first 3 elements of the array stdt[] with values known at compile-time, and then the next 3 students' information will be populated from user input. When I compile I get the error:

lab7.c: In function ‘main’:

lab7.c:16:9: error: expected expression before ‘{’ token
 stdt[0]={"John","Bishop","s1234","Inf",'m',18};
         ^

lab7.c:17:9: error: expected expression before ‘{’ token
 stdt[1]={"Lady","Cook","s2345","Eng",'f',21};
         ^

lab7.c:18:9: error: expected expression before ‘{’ token
 stdt[2]={"James","Jackson","s33456","Eng",'m',17};
         ^

How can I do this correctly?

Here is the code so far:

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

typedef struct {
    char *name;
    char *surname;
    char *UUN;
    char *department;
    char gender;
    int age;
} student_t;

int main() {
    int i;
    student_t stdt[6];
    stdt[0]={"John","Bishop","s1234","Inf",'m',18};
    stdt[1]={"Lady","Cook","s2345","Eng",'f',21};
    stdt[2]={"James","Jackson","s33456","Eng",'m',17};

    for(i=3;i<6;i++) {
        printf("First name: \n");
        scanf("%s",stdt[i].name);
        printf("Last name: \n");
        scanf("%s",stdt[i].surname);
        printf("UUN: \n");
        scanf("%s",stdt[i].UUN);
        printf("Department: \n");
        scanf("%s",stdt[i].department);
        printf("Gender (m/f): \n");
        scanf("%c",stdt[i].gender);
        printf("Age: \n");
        scanf("%d",stdt[i].age);
    }
    return 0;
}

You should check the return value of each of the scanf() calls to assure a value was actually input. — user3629249
– user3629249, Commented Nov 1, 2014 at 1:33
each of the scanf() format strings should include a leading space ' ' so unconsumed white space (like '\n') are consumed, otherwise scanf() can get hungup and not actually input some value. — user3629249
– user3629249, Commented Nov 1, 2014 at 1:35

Crowman · Accepted Answer · 2014-11-01 01:41:25Z

8

You're not "initializing" if you don't do it at the point of creation. You can do:

student_t stdt[2] = { {"John", "Bishop", "s1234", "Inf", 'm', 18},
                      {"Lady", "Cook", "s2345", "Eng", 'f', 21}
                    };

for as many as you have.

It's OK to not explicitly provide values for each member of the array. For the ones you don't explicitly initialize, pointer members will be implicitly initialized to NULL, and numeric members will be implicitly initialized to 0. In other words, this:

student_t stdt[4] = { {"John", "Bishop", "s1234", "Inf", 'm', 18},
                      {"Lady", "Cook", "s2345", "Eng", 'f', 21}
                    };

is equivalent to this:

student_t stdt[4] = { {"John", "Bishop", "s1234", "Inf", 'm', 18},
                      {"Lady", "Cook", "s2345", "Eng", 'f', 21},
                      {NULL, NULL, NULL, NULL, 0, 0},
                      {NULL, NULL, NULL, NULL, 0, 0}
                    };

For the curious, these rules derive from the C standard as follows.

From C11 Section 6.7.9.21:

If there are fewer initializers in a brace-enclosed list than there are elements or members of an aggregate, or fewer characters in a string literal used to initialize an array of known size than there are elements in the array, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration.

and for "the same as objects that have static storage duration" we have Section 6.7.9.10:

If an object that has automatic storage duration is not initialized explicitly, its value is indeterminate. If an object that has static or thread storage duration is not initialized explicitly, then:

if it has pointer type, it is initialized to a null pointer;

if it has arithmetic type, it is initialized to (positive or unsigned) zero;

if it is an aggregate, every member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

if it is a union, the first named member is initialized (recursively) according to these rules, and any padding is initialized to zero bits;

A struct is an "aggregate" in the sense of the third bullet above.

edited Nov 1, 2014 at 1:41

answered Nov 1, 2014 at 1:16

Crowman

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2 Comments

pixelfigure Over a year ago

So I can say stdt[6] while only telling it 3 elements and then I can still put the next 3 in with user input? Since it would appoint it the necessary slots of memory, right?

Crowman Over a year ago

Yes, that would be fine, you don't have to explicitly provide values for them all.

Jonathan Leffler · Accepted Answer · 2014-11-01 02:12:12Z

2

You can do almost what you wrote if you use C99 'compound literals':

stdt[0] = (student_t){ "John",  "Bishop",  "s1234",  "Inf", 'm', 18 };
stdt[1] = (student_t){ "Lady",  "Cook",    "s2345",  "Eng", 'f', 21 };
stdt[2] = (student_t){ "James", "Jackson", "s33456", "Eng", 'm', 17 };

However, this is assigning values to the array, not initializing the array. Further, since the array is local to main(), the values in the other three elements of the array are indeterminate; you cannot assume anything about them.

answered Nov 1, 2014 at 2:12

Jonathan Leffler

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