I have this "simple" problem: I have in input 2 int numbers and i must output them in decreasing order.
#include <stdio.h>
#include <iostream>
int fnum()
{
int NUM;
scanf("%d",&NUM);
return NUM;
}
void frisultato(int x,int y)
{
if (x>y)
{
printf("%d",x);
printf("%d",y);
}
else
{
printf("%d",y);
printf("%d",x);
}
return;
}
int main()
{
int A,B;
A=fnum;
B=fnum;
frisultato(A,B);
}
I recieve an error at
A=fnum;
B=fnum;
my compiler says: invalid conversion from int(*)() to int.
This is the first time i use functions, what is the problem? Thank you!
Michelangelo.
A=fnum;
B=fnum;
You're not actually calling the function fnum here. You're attempting to assign a pointer to the function to the int variables A and B.
To call the function, do this:
A=fnum();
B=fnum();
Sorry, but since you seem to be new at programming, I couldn't help but refactor/comment on your code:
#include <stdio.h>
#include <iostream>
int fnum()
{
int num;
scanf("%d",&num);
return num;
}
void frisultato(int x, int y)
{
if (x>y)
{
printf("%d",x);
printf("%d",y);
}
else
{
printf("%d",y);
printf("%d",x);
}
/* No need to return in void */
}
int main()
{
/*
Variables in C are all lowercase.
UPPER_CASE is usually used for macros and preprocessor directives
such as
#define PI 3.14
*/
int a, b;
a = fnum(); //Function calls always need parenthesis, even if they are empty
b = fnum();
frisultato(a, b);
/*
Your main function should return an integer letting whoever
ran it know if it was successful or not.
0 means everything went well, anything else means something went wrong.
*/
return 0;
}
Also, don't sign your name on StackOverflow questions.
Related
#include<stdio.h>
int fun()
{
static int num=16;
return num--;
}
void main()
{
for(fun();fun();fun())
printf("%d\n", fun());
}
Here what is meaning of for(fun();fun();fun()) ?
All I know about for loop is that for(initialization;condition;change in variable ) and they(ini..,cond...,chang...) should contain (some variable with value, algebraic condition, ..).
Please correct me.
Your code is equivalent to:
#include <stdio.h>
int fun()
{
static int num=16;
return num--;
}
void main()
{
fun();
while (fun())
{
printf("%d\n", fun());
fun();
}
}
Is it clear now?
I'm learning C because I have a book though it is as old as I am. I am using turboc++ and it works when I return 0; at the end of functions but I'm led to believe this was not always necessary. I am going through the exercises on functions.
outfloat will give me type mismatch in re-declaration of outfloat but the others work. What am I doing wrong?
#include "stdio.h"
main()
{
outchar('A');
outnum(2);
outfloat(3.3);
return 0;
}
outchar(char ch)
{
printf("%c",ch);
return 0;
}
outnum(int x)
{
printf("%d",x);
return 0;
}
outfloat(float z)
{
printf("%f",z);
return 0;
}
Since you don't provide a prototype for any of the functions used, the compiler assumes their parameters (and return type) are int.
For the first two functions that assumption "works".
For the last function, the assumption fails because the the parameter of type float is not compatible with (the assumed) type int.
Your best bet is to always provide prototypes for the functions you use. In the simple code you have you can use the fact that a function definition also serves as a prototype and move the definitions to before main().
#include <stdio.h>
/* prototypes */
int outchar(char ch);
int outnum(int x);
int outfloat(float z);
int main(void)
{
outchar('A');
outnum(2);
outfloat(3.3);
return 0;
}
int outchar(char ch)
{
printf("%c",ch);
return 0;
}
int outnum(int x)
{
printf("%d",x);
return 0;
}
int outfloat(float z)
{
printf("%f",z);
return 0;
}
I'm trying to create two functions. The first function accepts integer inputs from the user until they are between 0 and 100. The seconds function display the validation to the stdOut.
However, it keeps giving me an error saying "Expected expression" when I call the function.
#include <stdio.h>
// function prototype for the function input
int input(int);
// function prototype for the function validate
int validate(int);
//main function
int main(void)
{
//calling the function input
input(int x)
//calling the function validate
validate(int y)
return 0;
}
// Function definition for input
int input(int a)
{
int r;
printf("Enter the int value of r\n");
scanf("%d",&r);
}
// Function definition for validate
int validate(int b)
{
int r;
if(r>= 0 && r<= 100)
printf("Valid number");
else
printf("Invalid");
}
There is at least one bug on almost every line of this program.
This is a standard problem for which there is a whole lot of incorrect advice out there (most importantly, only the strtol/strtoul/strtod family of functions should be used to convert strings to numbers; never use the atoi family and never use scanf) so I am going to give a complete worked example of how to write this program correctly, including proper use of comments.
#include <stdio.h>
#include <stdlib.h>
#include <errno.h>
long read_number_in_range(const char *prompt, long lo, long hi)
{
// a signed 64-bit number fits in 21 characters, +1 for '\n', +1 for NUL
char buf[23], *endp;
long rv;
for (;;) {
puts(prompt);
if (!fgets(buf, sizeof buf, stdin)) {
perror("stdin");
exit(1);
}
errno = 0;
rv = strtol(buf, &endp, 10);
if (endp != buf && (*endp == '\0' || *endp == '\n')
&& !errno && rv >= lo && rv <= hi) {
return rv;
}
// if we get here, fgets might not have read the whole line;
// drain any remainder
if (!strchr(buf, '\n')) {
int c;
do c = getchar();
while (c != EOF && c != '\n');
}
puts("?Redo from start");
}
}
int main(void)
{
long val = read_number_in_range("Enter the int value of r", 0, 100);
// do something with val here
return 0;
}
Read on for line-by-line nitpicking of the original program.
#include <stdio.h>
Correct.
// function prototype for the function input
Comment redundant with code.
int input(int);
Function signature incorrect (see comments on body of function).
// function prototype for the function validate
Comment redundant with code.
int validate(int);
Function signature incorrect (see comments on body of function).
//main function
Comment redundant with code.
int main(void)
{
Correct.
//calling the function input
Comment redundant with code.
input(int x)
Variables cannot be declared inside function call expressions.
Return value of function is ignored.
Missing semicolon at end of line.
//calling the function validate
Comment redundant with code.
validate(int y)
Value returned from input should be passed to validate, presumably, instead of a new uninitialized variable.
Variables cannot be declared inside function call expressions.
Return value of function is ignored.
Missing semicolon at end of line.
return 0;
}
Correct.
// Function definition for input
Comment redundant with code.
int input(int a)
{
Parameter a is unnecessary.
int r;
Correct.
printf("Enter the int value of r\n");
Minor: use puts when there is nothing to format.
scanf("%d",&r);
Never use scanf.
}
Missing return r;.
// Function definition for validate
Comment redundant with code.
int validate(int b)
{
Function has no return value, so should be void validate(int b).
int r;
Unnecessary variable.
if(r>= 0 && r<= 100)
r should be b on this line.
printf("Valid number");
else
printf("Invalid");
Minor: again, puts.
}
Correct.
You have some stray ints in your calls, they need to go.
The calls should probably be:
x = input();
validate(x);
You can't pass an integer to a function and expect it to change in the caller's context, that is not how C's pass-by-value semantics work. You should just return the number from input() instead, i.e. its prototype should be int input(void);.
You need to add semicolons at the end of each line of code in your main() function, and also remove the type specifier in the function calls. Also, don't forget to declare the variables x and y somewhere:
int main(void)
{
int x=0;
int y=0;
//calling the function input
input(x);
//calling the function validate
validate(y);
return 0;
}
I bet the compiler asks you for an expression in some place where you write a function declaration with a missing semicolon.
//calling the function input
input(int x)
//calling the function validate
validate(int y)
Neither of these is a function call.
#include <stdio.h>
int input(int*);
int validate(int);
int main(void){
int x;
input(&x);
if(validate(x))
printf("Valid number");
else
printf("Invalid");
return 0;
}
int input(int *r){
printf("Enter the int value of r\n");
scanf("%d", r);
return *r;
}
int validate(int r){
return r>= 0 && r<= 100;
}
I am a Beginner in Functions.I wanna integrate that Functions in my main program. The program should scan a int number and then square it(sum=b*b). Then the program should output the result.
#include <stdio.h>
#include <stdlib.h>
void funktion(int);
void out(int);
int main(int)
{
int sum,b,v,w,z;
{
funktion();
calculator();
out();
printf("%i",sum);
}
return 0;
}
void funktion(int v)
{
printf("Enter any number that is to be squared!");
}
void calculator(int w) //calculate b*b
{
scanf("%i",&b);
sum=b*b;
}
void out(int z)
{
printf("Sum:");
}
Please give me some tips. ;)
Thx & Best regards!
Are you looking for this kind of solution: i just made little changes. you may need to do some initializing things if u gonna use this code
#include <stdio.h>
#include <stdlib.h>
void funktion(int);
void out(int);
int main()
{
int sum,b;
{
printf("Enter any number that is to be squared!");
scanf("%i",&b);
sum = funktion(b);
out(sum);
}
return 0;
}
int funktion(int b)
{
return b*b;
}
void out(int sum)
{
printf("Sum:%d",sum);
}
void funktion(int);
void out(int);
These two functions need not contain any arguments as you declared to be int's, because you are not performing any computation on sum, since they are just prints inside those functions.
So make them as,
void funktion(void);
void out(void);
and call those routines as,
funktion();
out();
void funktion(int)
{
printf("Enter any number that is to be squared!");
}
void out(int)
{
printf("Sum:");
}
to call these functions you are passing some value so you receive them with parameters.
void funktion(int a)
{
printf("Enter any number that is to be squared!");
}
void out(int b)
{
printf("Sum:");
}
You can make use of the passed value in functions using these variables, a nd b in corresponding functions.
If you dont want to access those passed values in the called functins then you dont need to pass the value. So change the functions to receive nothing, it can be
#include <stdio.h>
#include <stdlib.h>
void funktion();
void out();
int main()
{
int sum,b;
{
funktion();
scanf("%i",&b);
sum=b*b;
out();
printf("%i",sum);
}
return 0;
}
void funktion()
{
printf("Enter any number that is to be squared!");
}
void out()
{
printf("Sum:");
}
make the call to function without passing arguments. you can even specify void as type to indicate that function doesn't take any parameters.
To calculate in other function and return the value use return statement.
int calculator() //calculate b*b
{
scanf("%i",&b);
sum=b*b;
return sum;
}
change the function call statement to,
sum=calculator();
When I compile the following C function / program I get errors like "missing ';' before 'type' 'remainder' : undeclared identifier" - what is wrong with this function?
#include <stdio.h>
void conversionTo(int number,int base) {
if(number==0)
return;
int remainder=number%base;
conversionTo((number/base),base);
if(remainder<10)
printf("%c",'0'+remainder);
else
printf("%c",'a'-10+remainder);
}
int main() {
conversionTo(int number,int base);
return 0;
}
I'm not a C expert, but from experience very long ago I believe you cannot declare variables in the middle of a function.
Also, it's unclear what you are trying to do with the function / print statements.
Try this:
#include <stdio.h>
void conversionTo(int number,int base) {
int remainder=number%base;
if(number==0)
return;
conversionTo((number/base),base);
if(remainder<10)
printf("%c",'0'+ remainder); // Through the way ASCII works that gives the ASCII rep
// of the remainder.
else
printf("%c",'a'-10+remainder); // Hex digits (A-F).
}
int main() {
conversionTo(/*Any number here*/10, /*any base number here*/2);
return 0;
}
You need to defines variables, then they can be used.
So this:
int main() {
conversionTo(int number,int base);
return 0;
}
should become this:
int main(void)
{
int number;
int base:
number = 47;
base = 11;
conversionTo(number, base);
return 0;
}
Also non C99 compliant compilers do not like having variables declared in the middle of a context:
void conversionTo(int number,int base) {
if(number==0)
return;
int remainder=number%base; /* this would fail. */
conversionTo((number/base),base);
To get around this open another context:
void conversionTo(int number,int base) {
if(number==0)
return;
{
int remainder=number%base;
conversionTo((number/base),base);
if(remainder<10)
printf("%c",'0'+remainder);
else
printf("%c",'a'-10+remainder);
}
}
You have to invoke your function with a value or variable, not a declaration:
conversionTo(123, 10); // using constant value
or
int number = 123, base = 10; // variable declaration
conversionTo(number, base); // using variable
You are calling your function wrong - pass the arguments this way:
conversionTo(2,2); // assuming you want to convert 2 to binary
or
int number = 123, base = 10;
conversionTo(number, base); // note this is not the same number and base as in the definition of your conversionTo function
Full code:
#include <stdio.h>
void conversionTo(int number,int base)
{
if(number==0)
return;
int remainder=number%base;
conversionTo((number/base),base);
if(remainder<10)
printf("%c",'0'+remainder);
else
printf("%c",'a'-10+remainder);
}
int main()
{
conversionTo(2,2); // assuming you want to convert 2 to binary
return 0;
}
There are 3 things when using functions:
Function declaration / prototype - the prototype of your function is:
void conversionTo(int ,int );
Function definition:
void conversionTo(int number,int base /* Parameter list*/)
{
// your functionality
}
Function call where you pass your arguments to your function:
conversionTo(2,2);
The statement conversionTo(int number,int base); simply re-declares it. Try this even this will compile:
int main()
{
printf("Hello World");
int main();
}
conversionTo(int number, int base)
is the syntax for declaring which parameters the function can take. To actually call the function, you need to omit the type (assuming you have variables of the respective name)
int number = 5;
int base = 10;
conversionTo(number, base); // <-- no int here!
Or you can use numbers directly:
conversionTo(5, 10);
Your function definition number and base in your function declaration void conversionTo(int number, int base) are the names that the values passed to it will have inside the function. So, if you call conversionTo(2,5), 2 will be seen inside the function as number, while 5 will be seen as base.
If you want to use variables instead of contants to call the function, you could do that:
int main()
{
int base = 2;
int number = 5;
conversionTo(base, number);
return 0;
}
In this confusing example, the variables base and value have value 2 and 5, respectively. But as you pass them to the function, the values inside it will be number = 2 and base = 5. This shows that those variables are actually different, despite of having the same name.
You should compile with $CC -std=c99, to enable declaring variables in the middle of a block.
(see section 6.8 in the specification)
The declaration "int remainder" must come before any statements in a block.
A declaration may have an initializer as you have here.
You could do:
void conversionTo(int number,int base) {
if (number > 0) {
int remainder...
}
}
since the function will not work with negative numbers.
To fix the other bugs in the routine:
void conversionTo(int number,int base)
{
if(number>=0&&base>0&&base<=36)
{
int remainder=number%base;
number/=base;
if(number>0)conversionTo(number,base);
printf("%c",(remainder<10)?'0'+remainder:'a'+remainder-10);
}
}
This will print a 0 if number is zero and only recurse if needed.