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i have noted they don't really work the same as normal arrays that simply write
int x[20]={0};

this will make a array with 20 0's after one another.
but if i try to do the same thing with struct's it goes south fast.
typedef struct FancyName {
     int x[20];
};

int _tmain(int argc, _tchar* argv[])
{
     FancyName Fancy;
     Fancy.x={0};
}

the following errors would be "cannot convert from 'int' to 'int[100]'".
and nomatter how i write the compiler will not execute it.
Fancy.x[]={0};
Fancy.x[0]={0};
Fancy.x={0};

so my question becomes how do i easely change all the values of a struct array to one value?
Posted

the easiest way is
C++
typedef struct FancyName {
     int x[20];
};

FancyName Fancy;

memset( Fancy, 0, sizeof(Fancy) );

Another way is to initialize a struct and memcopy to another
C++
FancyName source;
source.x[0] = 3; //for illustration
FancyName Fancy;

memcpy( Fancy, source, sizeof(Fancy) );

The last way is stupid looping in arrays of struxts.

your struct is lagging of sense, because it is an array. structs making more sense like here:
C++
typedef struct ColorPoint{
 int x; //coordinate
 int y; //coordinate
 int z; //coordinate
 int color; //color value
 float alpha;//alpha value

ColorPoint colorPoint;
colorPoint.x = 1;
colorPoint.y = 2;
colorPoint.z = 3;
colorPoint.color = 0xff;//red
colorPoint.alpha = 0.5; //half transparent
 
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Quote:
typedef struct FancyName {
int x[20];
};
Using the C++ programming language, you don't need the typedef.



You might perform the inizialization to zero this way
C++
FancyName fc = {{0}};


or, using a modern C++ compiler
C++
FancyName fc{};

This would default-initialize all the members of the struct (see C++11 - the new ISO C++ standard: "uniform initialization"[^]).
 
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