Function with a pointer to an array argument in C

Question

I have written code below that contains a function with a pointer to a double array of size 3.My problems is this: when I pass the address of pointer to a double variable (that clearly isn't an array) and then want to change the value of this double variable in function "f" as written below, when I implement in this manner, the result is correct and the value of the variable is changed:

#include <stdio.h>
void f(double (*)[3]);
double a = 7.5;

int main()
{
   double* b = &a;
   f(&b);
   printf("a = %lf\n", a);
   return 0;

}
void f(double (*hi)[3])
{
   double **sth = (double **) hi;\
   *(*sth) = 1;

}

But when I implement as below the value isn't changed:

void f(double (*hi)[3]){
    (*hi)[0] = 1;
}

Any idea and suggestion is surely appreciated.

If you use pointers to arrays, you are almost certainly doing it wrong. In C, you pass a pointer to the first element of an array to functions, not a pointer to an array. — fuz
– fuz, Commented Oct 10, 2015 at 20:04
Also, converting a pointer to an array to a pointer to pointer and then dereferencing the result is undefined behaviour. Don't do that. — fuz
– fuz, Commented Oct 10, 2015 at 20:06
A pointer to array and a pointer to pointer are not the same thing. In addition you switch off compiler warnings by casting a spell. Don't do that, good C code works without casts. The suggestion would be, don't use casts, sort out your types. — Jens Gustedt
– Jens Gustedt, Commented Oct 10, 2015 at 20:07
Looks like you'll have to work on your understanding on arrays and pointers -- therefore please update your question. To get your code to consider this: (1) change f(&b); to f(b); .. (as you want to pass the pointer, not the address of the pointer) (2) change double (*hi)[3] to double (*hi). — harre
– harre, Commented Oct 10, 2015 at 20:39

noman pouigt · Accepted Answer · 2015-10-12 19:09:04Z

1

First fix the compilation errors your program is giving. After fixing those you will know the problem.

prog.c: In function 'main':
prog.c:8:6: error: passing argument 1 of 'f' from incompatible pointer type [-Werror=incompatible-pointer-types]
    f(&b);
      ^
prog.c:2:6: note: expected 'double (*)[3]' but argument is of type 'double **'
 void f(double (*)[3]);

http://ideone.com/mueyCU

#include <stdio.h>
void f(double (*)[3]);
double a = 7.5;

int main()
{
    double* b = &a;
    f(b);
    printf("a = %lf\n", a);
    return 0;

}

void f(double (*hi)[3]){
    (*hi)[0] = 1;
}

Above code is not the right way to do things and should be avoided at all costs.

edited Oct 12, 2015 at 19:09

answered Oct 12, 2015 at 17:58

noman pouigt

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7 Comments

Mehdi Over a year ago

actually I want to understand the syntax I have written above! I know it is complicated and there are much better ways than my code but I do not want the result with another code!!!

noman pouigt Over a year ago

@Mehdi: do you know what is this double (*hi)[3]?

Mehdi Over a year ago

I think this is a pointer to an array of double type @noman pouigt

noman pouigt Over a year ago

@Mehdi: I have edited the answer.Have a look and regarding your answer it is not right. double (*hi)[3] is a pointer to an array[3] of type double.

Mehdi Over a year ago

yes it works! but logicly we should send &b to function f, isn't it? @noman pouigt

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