Function and their types

[vead]

I have read it about function and their types in c programming. so after reading I tried to write program for function with no argument and no return type

This is my program ( void function_name (void)

#include<stdio.h>

void delay (void);

int main (void)
{
  int number;
  number = delay();
 
  printf(" Print number : %d",number);
 
   return 0;
}

int delay (void)
{
    unsigned int i;
    for (i = 0; i < 6; i++);
       {
           }     
}

error: void value not ignored as it ought to be
number = delay();

I don't know weather my example is suitable for this reason but I thought it would be best so I just wrote my program. But when I compiled program it's showing error. What's the wrong my idea or program

 

[(*steve*)]

Check the syntax of the for statement.

Also, if you have a function that doesn't return a value, why are you trying to assign a variable the return from it?

What do you expect it to do?

 

[vead]

Check the syntax of the for statement.

What do you expect it to do?

I studied about Function with no argument and no Return value and then tried to make my own program
I want to do like this example https://www.tutorialgateway.org/types-of-functions-in-c/

 

[(*steve*)]

You didn't study it very well.

Why are you expecting a return value from your function "delay"?

 

[vead]

You didn't study it very well.

Why are you expecting a return value from your function "delay"?

right I have done mistake i posted wrong program

#include<stdio.h>

void delay (void);

int main (void)
{
  delay();
 
   return 0;
}

void delay (void)
{
    unsigned int i;
    for (i = 0; i < 6; i++);
       {
           }    
}

I was trying to assign value of delay function to another variable and then wanted to know the the value of variable like this . but when I do, this I get error

int main (void)
{
  int number;
  number = delay();
 
  printf(" Print number : %d",number);
 
   return 0;
}

error: void value not ignored as it ought to be
number = delay();

 

[(*steve*)]

Please see post #2

 

[vead]

Please see post #2

Function with no argument and no Return value

#include<stdio.h>

void delay (void)
{
    unsigned int i;
    for (i = 0; i < 3 ; i++)
       {
            printf(" Print number : %d \n ",i);
         }     
}
int main (void)
{
   delay();
   return 0;
}

Print number : 0
Print number : 1
Print number : 2

Function with argument and no Return value

#include<stdio.h>

void delay (unsigned int number)
{
    unsigned int i;
    for (i = 0; i < number; i++)
       {
            printf(" Print number : %d \n ",i);
         }     
}
int main (void)
{
   delay(5);
   return 0;
}

Print number : 0
Print number : 1
Print number : 2
Print number : 3
Print number : 4

How to write program with no argument and Return value ?
How to write program with argument and Return value ?

 

[(*steve*)]

Now you seen to have done it.

 

[BobK]

I cannot even imagine what it is you are not understanding. A function with no return value does not have a return value. So we call it like this:

void delay();
...
delay();

If it had a return value, we could assign to to a variable of the same type:

int delay();
...
int x;
x = delay();

You last example did both of these correctly, so what is it you do not understand?

Bob

 

[vead]

I cannot even imagine what it is you are not understanding. A function with no return value does not have a return value.

Function with return value but no argument

#include<stdio.h>

int delay (void)
{
    unsigned int i;
    for (i = 0; i < 5; i++)
       {
            printf(" Print number : %d \n ",i);
         }  
     
         return i;
}
int main (void)
{
   int x;
   x = delay();
   printf(" Print delay : %d \n ",x);
   return 0;
}

Print number : 0
Print number : 1
Print number : 2
Print number : 3
Print number : 4

Print delay : 5

Look at following program

#include<stdio.h>

int delay (void)
{
    unsigned int i;
    for (i = 0; i < 5; i++)
       {
            printf(" Print number : %d \n ",i);
         }  
     
         return i;
}
int main (void)
{
 
   delay();
 
   return 0;
}

what happen when main function call to delay function, does it return numbers from o to 4 in main function or it return number 5 in main function?

i think , when we call delay function from main function then it will return 5 in main function.

 

[(*steve*)]

i think , when we call delay function from main function then it will return 5 in main function.

Neither, it returns nothing.

 

[vead]

Neither, it returns nothing.

I am confuse, can we say that below function return only integer value ( 0,1,2,3,4)

int delay (void)
{
    unsigned int i;
    for (i = 0; i < 5; i++)
       {
            printf(" Print number : %d \n ",i);
         }
  
         return i;
}

so when this function execute it return only integer value. it does not pass any argument , so when we call it from main function it doesn't return anything right ?

 

[(*steve*)]

No, it will return 5.

The arguments are separate from the return value.

 

[vead]

No, it will return 5.

The arguments are separate from the return value.

so does below program return only number 4

int delay (void)
{
    unsigned int i;
    for (i = 0; i < 4; i++)
       {
            printf(" Print number : %d \n ",i);
         }
 
         return i;
}

Look at another example. i tried to make program function with return value but no argument

#include<stdio.h>
int main(void)
{
    count_value();
   
    return 0;
}

unsigned int count_value (void)
{
  unsigned int value = 2;
  if ( value == 2)
    {
      print (" return value %d ",value);
       return 2;
    }    
 
  else
   {
      print (" return value %d ",value);
       return 1;
   }
       
}

It show some errors


warning: implicit declaration of function 'count_value' [-Wimplicit-function-declaration]
count_value();
^~~~~~~~~~~
hello.c: At top level:
hello.c:9:14: error: conflicting types for 'count_value'
unsigned int count_value (void)
^~~~~~~~~~~
hello.c:4:2: note: previous implicit declaration of 'count_value' was here
count_value();
^~~~~~~~~~~
hello.c: In function 'count_value':
hello.c:14:4: warning: implicit declaration of function 'print' [-Wimplicit-function-declaration]
print (" return value %d ",value);
^~~~~

 

[Harald Kapp]

warning: implicit declaration of function 'count_value' [-Wimplicit-function-declaration]

When you call count_value within main it has not yet been declared. So the compiler doesn't know this function. Compare with teh code on your post #10, where you declare delay() before calling it within main.
If you want to write the function after the calling code, you need to first declare the function to make it known to the compiler.

conflicting types for 'count_value'

This is a consecutive error due to the previous one. As the function declaration for count_value was missing when the compiler first encountered the call to count_value within main, the compiler assigned an implicit return type "int" to the function. When it later encounters the code for count_value, the definition as "unsigned int count_value (void)" conflicts with the implicit type int.

You need to delcare count_value() before calling it to make these errors disappear.

 

[vead]

You need to delcare count_value() before calling it to make these errors disappear.

Thanks. Does this program make any sense or it could be better

#include<stdio.h>
unsigned int count_value(void);
int main(void)
{
    count_value();
 
    return 0;
}

unsigned int count_value (void)
{
  unsigned int value = 1;
  if ( value == 2)
    {
      printf (" return value %d ",value);
       return value;
    }  
 
  else
   {
      printf (" return value %d ",value);
       return value;
   }
     
}

return value 1

Function with return value and argument

#include<stdio.h>

int delay (unsigned int number);

int main (void)
{
   delay(4);
   return 0;
}

int delay (unsigned int number)
{
    unsigned int i;
  
    for (i = 0; i < number; i++)
       {
            printf(" Print number : %d \n ",i);
       }
  
  
      return i;
}

Print number : 0
Print number : 1
Print number : 2
Print number : 3

if you see my I tried to write my own program's

1. Function with no return value and no argument
   
2. Function with no return value and argument
3. Function with return value and no argument
4. Function with return value and argument

I hope my all four program's are correct.

I just want to make sure my all program's are correct, are they correct ?

 

[BobK]

I think I see what your problem is with functions with a return value and no arguments. You are trying to make it return something different each time it is called, but not succeeding. Here it how it can be done:

int count = 0;

int counter()
{
    count++;
    return count;
}

int main()
{
    while (1)
    {
        printf("%d\n", counter());
    }
}

output:

1
2
3
4
...

The trick is that the function "counter()" has no arguments, but it uses a global variable, declared outside the function to decide what to return.

Bob

 

[vead]

I think I see what your problem is with functions with a return value and no arguments.

The trick is that the function "counter()" has no arguments, but it uses a global variable, declared outside the function to decide what to return.

I compiled your program

#include<stdio.h>

int count = 0;

int counter(void)
{
    count++;
    return count;
}

int main(void)
{
    while (1)
    {
        printf(" count %d \n", count);
    }
}

result :The system cannot execute the specified program.

I think we only print the value of variable not the function, that's why I wrote printf(" count %d \n", count); in place of your statement printf("%d\n", counter());

 

[BobK]

You can have any expression as an argument in the printf statement.

You can replace "count" in your printf with "counter()" and it will work correctly.

The way you have changed it, the function is not even being called.

Bob

 

[vead]

You can replace "count" in your printf with "counter()" and it will work correctly.

Sorry but it's not working on winGw. what's wrong

#include<stdio.h>

int count = 0;

int counter(void)
{
    count++;
    return count;
}

int main(void)
{
    counter();
 
   while (1)
    {
        printf(" count %d \n", counter());
    }
    return 0;
}

The system cannot execute the specified program