SUMMER TRAINING Free Tutorials  Go To Your University  Placement Preparation 
Project Based Best Summer Training Courses in Jaipur
Join our Telegram Channel To take free Online Courses
0 like 0 dislike
61 views
in Tutorial & Interview questions by Goeduhub's Expert (8.2k points)

Passing multidimensional arrays to a function

1 Answer

0 like 0 dislike
by Goeduhub's Expert (8.2k points)
 
Best answer
Multidimensional arrays follow the same rules as single-dimensional arrays when passing them to a function. However the combination of decay-to-pointer, operator precedence, and the two different ways to declare a multidimensional array (array of arrays vs array of pointers) may make the declaration of such functions nonintuitive. The following example shows the correct ways to pass multidimensional arrays.

#include <assert.h>

#include <stdlib.h>

/* When passing a multidimensional array (i.e. an array of arrays) to a   function, it decays into a pointer to the first element as usual.  But only   the top level decays, so what is passed is a pointer to an array of some fixed   size (4 in this case). */

void f(int x[][4])

{    

assert(sizeof(*x) == sizeof(int) * 4);

}

/* This prototype is equivalent to f(int x[][4]).   The parentheses around *x are required because [index] has a higher   precedence than *expr, thus int *x[4] would normally be equivalent to int *(x[4]), i.e. an array of 4 pointers to int.  But if it's declared as a   function parameter, it decays into a pointer and becomes int **x,   which is not compatable with x[2][4]. */

void g(int (*x)[4])

{    

assert(sizeof(*x) == sizeof(int) * 4);

}

/* An array of pointers may be passed to this, since it'll decay into a pointer   to pointer, but an array of arrays may not. */

void h(int **x)

{    

assert(sizeof(*x) == sizeof(int*));

}

int main(void)

{    

int foo[2][4];    

f(foo);    

g(foo);

    /* Here we're dynamically creating an array of pointers.  Note that the       size of each dimension is not part of the datatype, and so the type       system just treats it as a pointer to pointer, not a pointer to array       or array of arrays. */    

int **bar = malloc(sizeof(*bar) * 2);    

assert(bar);    

for (size_t i = 0; i < 2; i++)

{        

bar[i] = malloc(sizeof(*bar[i]) * 4);        

assert(bar[i]);    

}

h(bar);       

for (size_t i = 0; i < 2; i++)

{        

free(bar[i]);    

}    

free(bar);

}

Our Mentors(For AI-ML)


Sharda Godara Chaudhary

Mrs. Sharda Godara Chaudhary

An alumna of MNIT-Jaipur and ACCENTURE, Pune

NISHA (IIT BHU)

Ms. Nisha

An alumna of IIT-BHU

Related questions

 Goeduhub:

About Us | Contact Us || Terms & Conditions | Privacy Policy || Youtube Channel || Telegram Channel © goeduhub.com Social::   |  | 
...