Struct array initialization

Question

     struct forces{
        double fo[1];
        double ft[1];
    };
    int main(void){
        struct forces frc;
        frc.fo[0]=6;
        frc.fo[1]=56;
        frc.ft[0]=567;
       printf("%.0lf\n",frc.fo[1]);
        return 0;
    }

Please assist. Why is my printout always frc.ft[0]?

Welcome to StackOverflow. Is this C or C++? You should tell us the language. That aside, in both languages arrays are 0-based and only include N elements. so int foo[5] has exactly 5 elements, foo[0] ... f[4]. foo[5] does not exist, and neither does frc.fo[1]. — Dave S
– Dave S, Commented Sep 11, 2018 at 0:10

Dave S · Accepted Answer · 2018-09-11 00:17:18Z

1

With your declaration, fo[1] does not exist and writing to it is undefined behavior for C and C++.

Change your declaration to this to make fo[1] a valid element

struct forces{
    double fo[2];
    double ft[1];
};

In C and C++ (and most similar languages) a declaration of foo[5] creates a 5-element array with valid index values of [0] ... [4]. int foo[1] only has a valid index foo[0]

With your code you see fo[1] in place of ft[0] because your compiler placed the two double variables together in memory with no gap (padding) between them.

edited Sep 11, 2018 at 0:17

answered Sep 11, 2018 at 0:13

Dave S

1,5201 gold badge17 silver badges20 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

Enganeer85 Over a year ago

Thank you. This c code. This is a common mistake I overlooked. I will give it a go.

Collectives™ on Stack Overflow

Struct array initialization

1 Answer 1

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related