• C

Simple C Code,

hi experts,

I wrote a code in C as shown below and its giving the following error.
Cannot convert int* to int
Type mismatch in parameter1 in call to Junk(int,int)
Cannot convert int* to int
Type mismatch in parameter2 in call to Junk(int,int)

could anyone explain to me why this error? pls explain in a simple way, newbie here>)
#include<stdio.h>
#include<conio.h>
void junk(int,int);
void main()
{
int i=5,j=2;
junk(&i,&j);
printf ("\n%d%d",i,j);
}
void junk(int *i,int *j)
{

*i=*i**i;
*j=*j**j;
}
really appreciate the help
thanks in advance
deep
LVL 3
deepthijiAsked:
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avizitCommented:
also your expressions look messy ,
it is advisable to use brackets  to make the meaning more clear

change

*i=*i**i;
*j=*j**j;

to

*i  =  (*i)  *  (*i);
*j  =  (*j)  *  (*j);

This way its more readable and easier to get the meaning at a glance
0
 
avizitCommented:
in your functrion prototype you have


void junk(int,int);

while for your function definition you have


void junk(int *i,int *j)
{

*i=*i**i;
*j=*j**j;
}


so t hey dont match



0
 
avizitCommented:
so you can just change the funtion prototype

from

void junk(int,int);

to

void junk(int *,int *);



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deepthijiAuthor Commented:
sorrry i did not get that, As i said i just started learnning C,
>>in your functrion prototype you have
>>void junk(int,int);

>>while for your function definition you have
>>void junk(int *i,int *j)

what is wrong, should i declare void junk(int*,int*) at the bginning?
please explain some more,


0
 
avizitCommented:
right ..


in the third line of your code it is the function prototype , here you tell the compiler what yuor function looks like i.,e what is its return type and what are its arguments

so you declare the function as

void junk(int,int);   i.e you say that the function named 'junk' retunrs no value and takes two integers as arguments.

but when you actuallky define the function you are defining it as

void junk(int *i,int *j)
{

*i=*i**i;
*j=*j**j;
}

i.e here you say that the function  returns  no value ( i.e void) and takes two pointer to integers (int *i, int *j)   as arguments

so you see your declaration and definition has a mismatch

thats why the error


>>>>
Cannot convert int* to int
Type mismatch in parameter1 in call to Junk(int,int)
Cannot convert int* to int
Type mismatch in parameter2 in call to Junk(int,int)
<<<<<


hence to correct it you should change the prototype to match the fucntion definition
so yuo can chnage the third line of your program

from

void junk(int,int);

to

void junk(int *,int *);




0
 
deepthijiAuthor Commented:
Got it, thanks a loot for the help,
thanks once again.
0
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