int main(){
int right[2][3] = {
{1,4,6}, {2,7,5}
}
....
calc(right);
}
int calc(int ** right){
printf("%i", right[0][0]);
}
I calc function that calculate some numbers based on a matrix, 开发者_C百科but I dont' know why i get seg fault when I access the variable right within the calc function. does any body know the solution?
edit: right now that is all it's doing at calc function. I have some calc stuff but it's all commented out trying to figure out how to access this variable.
Two-dimensional arrays in C don't work the way you think they do. (Don't worry, you're not alone -- this is a common misconception.)
The assumption implicit in the code is that right
is an array of int *
pointers, each of which points to an array of int
. It could be done this way -- and, confusingly, the syntax for accessing such an array would be the same, which is probably what causes this misconception.
What C actually does is to make right
an array of 12 int
s, layed out contiguously in memory. An array access like this
a=right[i][j];
is effectively equivalent to this:
int *right_one_dimensional=(int *)right;
a=right[i*3 + j];
To pass your array to the calc
function, you need to do this:
int calc(int *right, size_t d){
// For example
a=right[i*d + j];
}
and then call it like this:
int right[2][3] = {
{1,4,6}, {2,7,5}
};
calc(&right[0][0], 3);
Edit: For more background on this, the question linked to in Binary Worrier's comment is definitely worth looking at.
Although a one-dimensional array is automatically converted to a pointer, the same does not hold for a multi-dimensional array and multi-level pointers.
If you change the order of the calc
and main
functions (or if you provide a prototype for calc
before main
), you will get a complaint from the compiler that it can convert right
to the type int**
.
The reason is that right
is declared as an "array of 4 arrays of 3 int
". This can be automatically converted to "pointer to array of 3 int
" (int (*)[3]
), but that is where the conversions stop.
calc
on the other hand expects a "pointer to a pointer to int
", which is a completely different beast from a "pointer to array of 3 int
".
There are two possible solutions:
- Change
calc
to accept a pointer to an array (or array of arrays):
int calc(int right[][3])
- Change
right
to be a pointer to a pointer:
int temp_array[4][3];
int* temp_array2[4] = { temp_array[0], temp_array[1], temp_array[2], temp_array[3] };
int** right = temp_array2;
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