how to initialize members of an array of struct in C

Question

I have the following struct:

typedef struct {
    int someArray[3][2];
    int someVar;
} myStruct;

If I create an array of this struct in my main (like the following), how would I initialize it?

int main() {
    myStruct foo[5];
}

I want to initialize the above array of struct in a way similar to initilazing a normal array (see below):

int main() {
    int someArray[5] = {1,4,0,8,2};
}

Just because some things can be done, doesn't necessarily mean you should do them. Consider how someone who had never seen your code before would perceive an initialization like that. — Hunter McMillen
– Hunter McMillen, Commented Mar 12, 2012 at 14:47
@HunterMcMillen - there are many cases where this is a better option than others. — MByD
– MByD, Commented Mar 12, 2012 at 14:57
@BinyaminSharet I just believe in terms of maintenance, it would be more clear to initialize each piece clearly commenting each step. Can you give an example situation where this would be preferred? None come to mind for me. — Hunter McMillen
– Hunter McMillen, Commented Mar 12, 2012 at 15:03
@HunterMcMillen - for example, when you need a big map, or dictionary. — MByD
– MByD, Commented Mar 12, 2012 at 15:12
@HunterMcMillen: One place is when the structure should be const; if you have the initializer, the compiler can enforce the const-ness by placing the initialized data in a read-only section of memory. Also, the intialization leads to more compact code compared with run-time initialization piece-meal. There might be a case for using designated initializers, though it is hardly necessary here if the type definition and the initializer are with easy viewing distance of each other. — Jonathan Leffler
– Jonathan Leffler, Commented Mar 12, 2012 at 15:20

John Bode · Accepted Answer · 2012-03-12 16:39:36Z

Work from the outside in. You know you have an array of 5 things to initialize:

mystruct foo[5] = { 
                    X, 
                    X, 
                    X, 
                    X, 
                    X 
                  };

where X is a stand-in for initializers of type mystruct. So now we need to figure out what each X looks like. Each instance of mystruct has two elements, somearray and somevar, so you know your initializer for X will be structured like

X = { Y, Z }

Substituting back into the original declaration, we now get

mystruct foo[5] = { 
                    { Y, Z }, 
                    { Y, Z }, 
                    { Y, Z }, 
                    { Y, Z }, 
                    { Y, Z } 
                  };

Now we need to figure out what each Y looks like. Y corresponds to an initializer for a 3x2 array of int. Again, we can work from the outside in. You have an initializer for a 3-element array:

Y = { A, A, A }

where each array element is a 2-element array of int:

A = { I, I }

Subsituting back into Y, we get

Y = { { I, I }, { I, I }, { I, I } }

Substituting that back into X, we get

X = { { { I, I }, { I, I }, { I, I } }, Z }

which now gives us

mystruct foo[5] = {
                    { { { I, I }, { I, I }, { I, I } }, Z },
                    { { { I, I }, { I, I }, { I, I } }, Z },
                    { { { I, I }, { I, I }, { I, I } }, Z },
                    { { { I, I }, { I, I }, { I, I } }, Z },
                    { { { I, I }, { I, I }, { I, I } }, Z }
                  };

Since Z is a stand-in for an integer, we don't need to break it down any further. Just replace the Is and Zs with actual integer values, and you're done:

mystruct foo[5] = {
                    {{{101, 102}, {201, 202}, {301, 302}}, 41},
                    {{{111, 112}, {211, 212}, {311, 312}}, 42},
                    {{{121, 122}, {221, 222}, {321, 322}}, 43},
                    {{{131, 132}, {231, 232}, {331, 332}}, 44},
                    {{{141, 142}, {241, 242}, {341, 342}}, 45}
                  };

Jonathan Leffler · Accepted Answer · 2012-03-12 14:47:46Z

3

Wrap the initializer for each structure element of the array in a set of braces:

myStruct foo[5] =
{
   { { { 11, 12 }, { 13, 14 }, { 55, 56 }, }, 70 },
   { { { 21, 22 }, { 23, 24 }, { 45, 66 }, }, 71 },
   { { { 31, 32 }, { 33, 34 }, { 35, 76 }, }, 72 },
   { { { 41, 42 }, { 43, 44 }, { 25, 86 }, }, 73 },
   { { { 51, 52 }, { 53, 54 }, { 15, 96 }, }, 74 },
};

answered Mar 12, 2012 at 14:47

Jonathan Leffler

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5 Comments

still.Learning Over a year ago

I think you might have missed the '{' on the second line. It's the same thing as your answer

Jonathan Leffler Over a year ago

@BinyaminSharet: GCC is happy with the code. There are a balanced set of open and close braces on each data line, with the surrounding { and }; at the beginnings of the lines before and after the data lines.

MByD Over a year ago

Indeed, I missed the first { sorry for that.

Ioan Over a year ago

Just being picky; you still have an extra comma on the last element. :-) It may not be a syntax error for GCC, but might elsewhere.

Jonathan Leffler Over a year ago

@loan: a trailing comma at the end of the initializers had better not be a syntax error anywhere; it would be a broken compiler that complained since the both the C89 and C99 standards mandate that it be allowed. In C89, §6.5.7 (C99 §6.7.9) Initialization, initializer: assignment-expression | { initializer-list } | { initializer-list , } explicitly calling out the trailing comma. C89 did not allow the comma at the end of an enumeration; C99 does. C2011 follows C99 AFAIK.

MByD · Accepted Answer · 2012-03-12 14:45:53Z

0

Like that:

int main() {
                      //   someArr initialization  | someVar initialization
    mystruct foo[5] = { { { {1, 2}, {1,2}, {1, 2} }, 1                      }, // foo[0] initialization
                       //...
                      };
}

answered Mar 12, 2012 at 14:45

MByD

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