c - Acessing a 2D array inside a function -

i have function accepts int* pinput[] argument.

void process(int*  pinput[], unsigned int num); 

i have call function via 2 methods as

main() { int *pin[2]; int input[2][100] = {0};  pin[0] = ( int* )malloc( 100 * sizeof( int) ); pin[1] = ( int* )malloc( 100 * sizeof( int) );  process( pin, 2 ); process(  ( int** )input, 2 ); } 

then how can access each value of pinput inside function 'process'?i cannot access directly pin[0][0].

how can access each value of pinput inside function 'process'?i cannot access directly pin[0][0].

no! can access way: pinput[0][0] if input pass pin. because pin array of int*s i.e. it's of type int *[n] each of element pointing array of ints. decay int**.

however, if want pass input, 2d array of ints, you've more since 2d array doesn't decay double pointer, t** pointer array, t (*) [n]. because array decay not recursive, happens first level. alternatively, can (live example)

pin[0] = input[0]; pin[1] = input[1]; 

and pass pin process. here pin surrogate input , needs have many elements input, not elegant solution. better way pass input, when know dimensions during compile-time is

void process(int (*pinput)[100], size_t rows) { } void process(int input [2][100], size_t rows) { } /* these 2 same; compiler never sees `2`. input's type int(*)[100] */ 

read on array decay understand situation better.

aside

• do cast result of malloc? no, not need cast return value of malloc in c.
• what should main() return in c , c++? return type of main should int.

related

• c-faq: my compiler complained when passed two-dimensional array function expecting pointer pointer.
• what array decaying?
• why can omit first dimension of multidimensional array when passing function