Initialize array of structs in C

Question

I am having a problem initializing an array of structs. I'm not sure if I am doing it right because I get "initialization from incompatible pointer type" & "assignment from incompatible pointer type". I added in the code where I get these warnings, and when I try to print the data from the struct I just get garbage such as @@###

typedef struct
{
    char* firstName;
    char* lastName;
    int day;
    int month;
    int year;
} student;

// Initialize array

student** students = malloc(sizeof(student));
int x;
for(x = 0; x < numStudents; x++)
{
    // Here I get: "assignment from incompatible pointer type"
    students[x] = (struct student*)malloc(sizeof(student));
}

int arrayIndex = 0;

// Add struct

// Create student struct

// Here I get: "initialization from incompatible pointer type"
student* newStudent = {"john", "smith", 1, 12, 1983};

// Add it to the array
students[arrayIndex] = newStudent;
arrayIndex++;

Your code seems to be half way between dynamically creating an array of student structs, and dynamically creating an array of pointers to student structs, and then dynamically creating each student struct which is pointed to. It's not obvious which one you are trying to do, which makes answering this difficult. — Pete Kirkham, Nov 13 '10 at 16:43

pmg · Accepted Answer · 2019-07-31T14:37:02.923

31

This is incorrect:

student** students = malloc(sizeof(student));

You do not want a **. You want a * and enough space for how ever many students you need

student *students = malloc(numStudents * sizeof *students); // or sizeof (student)
for (x = 0; x < numStudents; x++)
{
    students[x].firstName = "John"; /* or malloc and strcpy */
    students[x].lastName = "Smith"; /* or malloc and strcpy */
    students[x].day = 1;
    students[x].month = 12;
    students[x].year = 1983;
}

If you still want to use the code in your "//add struct" section, you'll need to change the line:

student* newStudent = {"john", "smith", 1, 12, 1983};

to

student newStudent = {"john", "smith", 1, 12, 1983};

You were getting "initialization from incompatible pointer type" because you were attempting to initialize a pointer to student with an object of type student.

edited Jul 31 '19 at 14:37

answered Nov 13 '10 at 16:42

pmg

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1

Surely you mean `student* students = malloc(numStudents * sizeof(student) )`, right? – GnP Jul 16 '16 at 23:38
3

@GnP: I prefer to use the object itself as a argument to the `sizeof` operator. With `struct something *ptr;` applying the operator to the (parenthesised) type `struct something` or to the object `*ptr` yields the same value. `sizeof *ptr == sizeof (struct something)`. – pmg Jul 17 '16 at 19:10
Thanks @prng. I was confused by the syntax until I realized `sizeof` is a unary operator and not a function in C. – GnP Jul 17 '16 at 20:32
why I cannot access my member variable using ``students[x]->firstName`` for instance? – xxks-kkk Jan 04 '17 at 02:31
@zack: because `.` is used with objects, and `->` is used with pointers. After the declaration `student *students;` every indexing into `students` yelds an object of type `student`. – pmg Jan 04 '17 at 09:30

score 8 · Answer 2 · answered Nov 13 '10 at 16:42

Unrelated to the compiler warnings, but your initial malloc is wrong; you want:

malloc(sizeof(student *)* numStudents)

To allocate room for a total of 'numStudents' pointers to a student. The line:

students[x] = (struct student*)malloc(sizeof(student));

Should be:

students[x] = (student*)malloc(sizeof(student));

There's no such thing as 'struct student'. You've declared an unnamed struct and typedef'd it to 'student'. Compare and contrast with:

struct student
{
    char* firstName;
    char* lastName;
    int day;
    int month;
    int year;

};

Which would create a 'struct student' type but require you (in C) to explicitly refer to struct student rather than merely student elsewhere. This rule is changed for C++, so your compiler may be a bit fuzzy about it.

As for:

student* newStudent = {"john", "smith", 1, 12, 1983};

That should be:

student newStudent = {"john", "smith", 1, 12, 1983};

As the curly brace syntax is a direct literal, not something somewhere else that you need to point to.

EDIT: on reflection, I think aaa may have taken more of an overview of this than I have. Is it possible that you're inadvertently using an extra level of pointer dereference everywhere? So you'd want:

student* students = malloc(sizeof(student) * numStudents);

/* no need for this stuff: */
/*int x;
for(x = 0; x < numStudents; x++)
{
    //here I get: "assignment from incompatible pointer type" 
    students[x] = (struct student*)malloc(sizeof(student));
}*/

int arrayIndex = 0;

And:

student newStudent = {"john", "smith", 1, 12, 1983};

//add it to the array
students[arrayIndex] = newStudent;
arrayIndex++;

Subject the array not being used outside of scope of newStudent. Otherwise copying the pointers to strings is incorrect.

score 2 · Answer 3 · answered Nov 13 '10 at 16:42

2

student* students = malloc(sizeof(student)*numStudents);
int x;
for(x = 0; x < numStudents; x++)
{
    student newStudent = {"john", "smith", 1, 12, 1983}; // string copy are wrong still
    students[x] = newStudent;
}

answered Nov 13 '10 at 16:42

Anycorn

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score -1 · Answer 4 · answered Nov 13 '10 at 16:43

-1

When initializing, shoudn't it be like this?

student** students = (struct student**)malloc(sizeof(student*)*numStudents);

However, why pointer to a pointer? Just with a pointer to struct is enough I think.

answered Nov 13 '10 at 16:43

frarees

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Initialize array of structs in C

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