Compiler error in C, Error : Expected Expression - c

Question

I have run into a problem I don't really know how to fix. I mostly code in Python, and this is my first program in C.
#include <stdio.h>
int ask(void) {
scanf("Var");
return 0;
}
int count(ask) {
scanf("number1");
return 0;
}
int main(void) {
printf("This is my first program!\n");
printf("I hope this program turns out well.");
printf("I don't really know what to do, but i think im progressing.\n");
printf("But yeah, This is my first program.\n");
printf("Type an Number");
ask();
count(ask());
printf("Thanks!");
printf(%count%);
return 0;
}
However, I keep getting an error.
main.c:22:10: error: expected expression
printf(%count%);
^
main.c:22:17: error: expected expression
printf(%count%);
^
2 errors generated.
compiler exit status 1
What I want it to do is, the user types a number, and then it prints out that number. It's not complete though. I want it to write numbers 1 - the users input, and when it gets the number right, it prints "Your number is:" (number)

Answer 1

The problem (as already pointed out) is that you aren't actually getting and storing the value from the scanf() call. Additionally, printf(%count%) is not valid C syntax. You need to use printf("%d", count).
Putting all that together:
#include <stdio.h>
int ask(void) {
int input_number;
scanf("%d", &input_number);
getchar(); # This is so that the '\n' in is read when you hit Enter
return input_number;
}
int main(void) {
printf("This is my first program!\n");
printf("I hope this program turns out well.");
printf("I don't really know what to do, but i think im progressing.\n");
printf("But yeah, This is my first program.\n");
printf("Type an Number");
int input_number = ask();
printf("Thanks!");
printf("The number you entered is %d\n", input_number);
return 0;
}
Some things to read to avoid making mistakes like these:
printf tutorial: https://www.tutorialspoint.com/c_standard_library/c_function_printf.htm
scanf tutorial: https://www.tutorialspoint.com/c_standard_library/c_function_scanf.htm

Answer 2

You are using incorrect format for printing.
you are using this code printf(%count%); instead you should use this code printf("%d",count(variablename));
Here is the code that you can use:
#include <stdio.h>
int ask(void) {
//scanf("Var");
int var; //declaring integer type varible in C
scanf("%d",&var); //taking input
return var; //returning what is being input by user
}
//int count(ask) //its function you are passing
//so it should have parenthesis
int count( int a )
{
return a;
}
int main(void) {
printf("This is my first program!\n");
printf("I hope this program turns out well.");
printf("I don't really know what to do, but i think im progressing.\n");
printf("But yeah, This is my first program.\n");
printf("Type an Number: ");
//ask();
count(ask()); // when you will call it will automatically run ask as well
// and it will pass int value that user will input
printf("Thanks!");
int var1=ask(); // this varible var1 will store value returned by ask()
printf("%d",count(var1) ); //this will print that value
//I dont know what you wanted to do here
//maybe you wanted to print value returned by count function
//so it can be done by this
return 0;
}
printf format: printf("%formatspecifier",variable name); Format specifiers are %d for a integer. %f for a floating value. %c for a charcater. s for a string and so on.

Answer 3

In C, we use % to specify output/input type. %d for integer, %f for float, %c for character, %s for string and thats your basic knowledge.
for printf:
printf("%d", varname);
for scanf:
scanf("%d", &varname);
'&' means location in memory.
Your program got many syntax errors, here is some code:
#include<stdio.h>
int ask(){
int varin;
scanf("%d", &varin);
return (varin);
}
int count(int countin){
return (countin); //just an example code, or whatever you wanna do here.
}
int main(){
int out;
printf("Whatever\n");
out = ask();
count(out);
printf("%d", out);
return 0;
}