This question already has answers here:
Difference between int main() and int main(void)?
(10 answers)
Closed 9 months ago.
#include <stdio.h>
int main()
{
static int i = 5;
if (--i){
printf("%d ", i);
main(10);
}
}
I was reading article on geeksforgeek the difference between int main(void) and int main() but I am confused. How main function taking arguement?
int main(void) { }
Here, you are telling your compiler explicitly that your main() function is not going to take any arguments. This prototype of main() function is standard.
int my_func(void)
{
return 100;
}
// the above function can be called as:
// 1. my_func(); --> good
// 2. my_func(21); --> error
// 3. my_func("STACKOVERFLOW"); --> error
int main() { }
Here, you are telling your compiler implicitly that your main() function is not going to take any arguments. But, there's a plot twist, in C leaving the function without any arguments (type function() { }) means you can put any value/variable in it while calling that function. This prototype should be avoided.
int my_func()
{
return 100;
}
// the above function can be called as:
// 1. my_func(); --> no error
// 2. my_func(21); --> no error
// 3. my_func("STACKOVERFLOW"); --> no error
Related
I am new to C programming and I am trying to create functions. The first function executes but the second one doesn't.
#include <stdio.h>
char get_char();
int main(void)
{
char ch;
printf("Enter a character > ");
scanf("%c", &ch);
return ch;
}
int get_int()
{
int i;
printf("Enter an integer between 0 and 127 > ");
scanf("%d", &i);
return i;
}
For anyone else that arrives here, you may solve your problem, by making sure your main function is at the bottom of the file.
Why? Because if function a calls function b, a should be before b.
main is the entry point for your program. The C environment calls main when your program is executed. get_int() is not the name for an entry point, so the fact that you never call it directly or indirectly in main means it will never be executed.
You also didn't declare it before main meaning your compiler will warn about not finding it, but since get_int returns int it will link successfully regardless.
Fix:
int get_int();
int main ()
{
//...
}
int get_int()
{
//...
}
Your second function isn't called and it should be as follows:
int main()
{
int m = 10;
get_int(m);
return 0;
}
int get(int num)
{
int multiply = num * num;
return multiply;
}
I have learned how to use functions and structs and pointers. I want to combined them all into one. But the code that I write doesn't seem to work. The compiler tells me the test is an undeclared identifier. Here is the code:
#include <stdio.h>
#include <stdlib.h>
struct character
{
int *power;
};
void test (use_power)
int main ()
{
test (use_power)
printf("%d\n",*power);
return 0;
}
void test ()
{
int use_power = 25;
struct character a;
a.power = &use_power;
}
Your code has many mistakes it can't even compile
Multiple missing semicolons.
Implicit declaration of test() here
test (use_power)
with a missing semicolon too.
power is not declared in main().
This line
void test use_power()
does not make sense and is invalid, and also has no semicolon.
The a instance in test() defined at the end is local to test() and as such will be deallocated when test() returns. The use_power int, has exactly the same problem and trying to extract it's address from the function is useless because you can't access it after the function has returned.
I have no idea what you were trying to do, but this might be?
#include <stdio.h>
#include <stdlib.h>
struct character {
int *power;
};
/* Decalre the function here, before calling it
* or perhaps move the definition here
*/
void test(struct character *pointer);
/* ^ please */
int
main(void) /* int main() is not really a valid signature */
{
struct character instance;
test(&instance);
if (instance.power == NULL)
return -1;
printf("%d\n", *instance.power);
free(instance.power);
return 0;
}
void
test(struct character *pointer)
{
pointer->power = malloc(sizeof(*pointer->power));
if (pointer->power != NULL)
*pointer->power = 25;
}
Your code seems to be wrong. Your definition for test contains no arguments as
void test ()
{
int use_power = 25;
struct character a;
a.power = &use_power;
}
but your prototype contains one argument
void test (use_power)
which is wrongly put. First there are no semicolons; at the end of your prototype declaration, secondly by looking at your code, use_power is a variable and not a datatype so it cannot be present solely in a function declaration.
You will get an argument mismatch error.
You have used the line in main()
printf("%d\n",*power);
which is absolutely wrong. you cannot access any member of a structure without a structure variable.
And again, you have not mentioned the; after your call to the incorrect test()before this line
As you have not put your question so properly, I must figure out what you wish to achieve. I bet you want to hold the address of a integer in the pointer member of a structure and then print its value.
Below is a code snippet which will work as you desire.
#include <stdio.h>
#include <stdlib.h>
struct character
{
int *power;
};
struct character a; //define a structure variable
void test ();
int main ()
{
test ();
printf("%d\n",*(a.power)); // print the member of structure variable a
return 0;
}
void test ()
{
int use_power = 25;
a.power = &use_power;
}
example
#include <stdio.h>
struct character {
int *power;
};
void test(struct character *var);
int main (void){
struct character use_power;
int power = 5;
use_power.power = &power;
test(&use_power);
printf("%d\n", power);
return 0;
}
void test(struct character *var){
int use_power = *var->power;
*var->power = use_power * use_power;
}
This question already has answers here:
C code explanation
(2 answers)
Closed 6 years ago.
int main(int argc, char **argv)
{
int (*func)();
func = (int (*)()) code;
(int)(*func)();
}
the variable code has some shellcode in it
Function pointers. This code snippet should help you understand.
#include <stdio.h>
int Hello();
int code();
int main(int argc, char **argv)
{
int (*func)(); //pointer to function that takes no arguments quivalent to: int (*func)(void);
func =&Hello;
int x = func();
printf("%d\n", x);
func = (int (*)()) code; // Assigns the pointer from the code function to the func pointer
x = code();
printf("%d", x);
}
int code()
{
printf("code returns: ");
return 500;
}
int Hello()
{
printf("hello returns: ");
return 1;
}
code probably is a variable that correspond to the address of some machine code in memory. Then a pointer to function that takes no parameter and returns an int is set to that address and the function is called. int f() is the prototype for a function with no param and int as return value, then int (*pf)() is a pointer to such a function.
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.
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.