called object 'strn' is not a function [closed] - c

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While compiling the below code, i am getting error
"called object strn is not a function"
tired of this error !! need a solution!!
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define num 400
int main()
{
char strn[num];
int count;
int a=0,e=0,i=0,o=0,u=0;
printf("enter your string!\n");
gets(strn);
for(count=0;count<strlen(strn);count++)
{
if ( strn(count)=='a' )
{
a++;
}
if (strn(count)=='e')
{
e++;
}

Answer 1

You are trying to use strn as if it's a function: strn(count).
You are presumably trying to access the value at the count index, so you should use strn[count].

Answer 2

The error is quite indicative. You have declared strn as an array of characters.
char strn[num];
And using it as strn(count) which is wrong. The compiler considers it as a function. You should use square brackets [ ] instead of parentheses ( ).

Answer 3

In your code, strn(count) represents a function call to strn() with one argument count. What you need is to use the Array subscripting operator [], not ()
You need to change
strn(count)
to
strn[count]
Also, please consider using fgets() over gets().

Answer 4

The subscript operator uses symbols [] for enclosing index. So for example instead of
strn(count)=='a'
you have to write
strn[count]=='a'
Also function gets is not supported any more by the C Standard because it is an unsafe function. Use instead fgets.
The program can look like
#include <stdio.h>
#include <ctype.h>
#define num 400
int main( void )
{
char strn[num];
char *p;
int a = 0, e = 0, i = 0, o = 0, u = 0;
printf( "Enter your string: " );
fgets( strn, num, stdin );
for ( p = strn; *p != '\0'; ++p )
{
char c = tolower( *p );
switch ( c )
{
case 'a':
a++;
break;
case 'e':
e++;
break;
// and so on...