I was doing this question and I have a doubt.
#include <stdio.h>
int main(void)
{
int fun (int);
int i=3;
fun(i=fun(fun(i)));
printf("%d\n",i);
return 0;
}
int fun ( int i )
{
i++;
return(i);
}
I have a doubt when it gets to
fun ( i = 5 )
What happens with this? Will the value of i go to 6 or it will be 5.
According to me, it should be 6. But that is not the correct answer.
In C, parameters are passed by value. The variable i in the main function is actually different from the i inside fun(), because its value is copied when it is passed into the function.
When you call i = fun(fun(i)), 5 is isassigned into i in the main function. However, the call to fun(5) that returns 6 does not assign its result back into i, leaving it unchanged. When the output is printed, i is still 5.
This is related to scope. In the function scope, variables defined in that scope or the parameters pass in does not any effect on outer scope variables, unless it is a pointer. So, the output of fun will be assigned to i in the fun(i = 5) but the internal operations of fun, do not effect the outer scope i. So it stays as it is before fun last call. The output is 5.
The result of the call to fun() is not assigned to i. Therefore 5 is expected, not 6.
Related
This is for programming in C.
Say I had the follow code in my program:
int fun1(int x);
int main (void)
{
int a = 5;
a = fun1(10);
}
int fun1(int x)
{
\\Program arbitrarily ends here
return x;
}
In my memory diagram what would the value of a be assuming the program terminates before fun1 is able to return a value? Would the value of a be undetermined (??) or would it be 5?
The value of a is already initialized to 5. Now, according to the condition, you want to know the results of that circumstance when the fun1() ends before it could return a value; assume the following:
int fun1(int x)
{
// return x;
}
Here we've supposed the function quits before returning the value.
You'll still get the output 5 before, during or after execution of the program since it returns nothing but the variable a is preassigned to 5 and it can't be changed to 10 unless the function returns 10 and assigns to the variable.
But remember, if you don't assign anything to a, then it may show an unexpected value (I got 4195638 when used printf() for a).
#include <stdio.h>
int main()
{
static int i = 5; // here
if (--i){
printf("%d ", i);
main();
}
}
Output: 4 3 2 1
#include <stdio.h>
int main()
{
int i = 5; // here
if (--i){
printf("%d ", i);
main();
}
}
Output: 4 4 4 4... (Segmentation fault)
Any idea how static int variable is taken into account only once and int is taken over and over again?
When you declare a static variable in a function, the function "remembers" the last value of the variable even after it terminates.
void Foo() {
static int x = 5;
}
In the example above, you are telling the compiler that x shall be "remembered" and has an initial value of 5. Subsequent calls to Foo() does not reassign x a value of 5, but uses the previously remembered value.
In contrast:
void Bar() {
int x = 5;
}
Here, you are telling the compiler that everytime Bar() executes, a new variable x is to be created on the stack and assigned a value of 5.
A variable declared as static inside of a function retains its value each time the function is called. It is initialized only once, when the function is first called. So when the function calls itself the value of i is retained from the prior call.
A local variable that is not static is specific to a given call of a function, so each time the function is called a new copy of the variable is created and initialized. This results in i being 5 each time, which in turn results in infinite recursion leading to a stack overflow and a core dump.
#include <stdio.h>
#include <stdlib.h>
int f(int x) {
return x;
}
int main ( int argc,char * argv[]) {
int a=4;
f(a);
printf("PASSED!\n");
return 0;
}
What happens when you call f(a) without assigning it to anything?
What happens when you call a function with return value without assigning it to any variable?
The return value of a function need not be used or assigned. It is ignored (usually quietly).
The function still executes and its side effects still occur.
Consider the 3 functions: int scanf(), int f(), and int printf(), their return values are all ignored yet the functions were still executed.
int a=4;
scanf("%d", &a);
f(a);
printf("PASSED!\n");
It is not good to ignore return values in robust code, especially scanf().
As commented by #Olaf, a warning may be enabled by some compilers.
Explicit ignoring the result of a function is sometime denoted with (void) to quiet that warning.
(void) f(a);
Using your example, we can look at how it evaluates line by line. Starting in main.
int a=4;
We now have a variable a with the value 4.
f(a);
So now the function f is called with a, which has a value of 4. So in the function f, the first parameter is named x and it just returns that parameter x.
So the evaluation of
f(a);
is just
4;
And a program like this compiles and runs perfectly fine.
int main(int argv, char *argv[]) {
1 + 1;
return 0;
}
What happens when you call f(a) without assigning it to anything?
--> Nothing at all.
What happens when you call a function (which has return value) without assigning it to anything?
-->The function will be executed, either make no sense like your case or make a lot of senses like modifying a static variable or a global variable. The return value will be ignored.
The return value will normally be stored in a register and will not fade.
It will be overwritten when the register is needed by the compiler.
If the function is inline it may be detected by the compiler that the value isn't used and ignore the value from being computed at all.
This question already has answers here:
What (actually) happens, when a function with the warning "control reaches end of non-void function" is called?
(7 answers)
Closed 7 years ago.
I am newbie. So please bear with me,
#include<stdio.h>
int abc(int k)
{
k++;
}
int main()
{
int a=1;
printf("%d",abc(a));
return 0;
}
Output of above program is : 1
My question is shouldn't the output should be '2' as the actual parameter is passing the value of '1' to the formal parameter and it has to be incremented by the function abc.
And when I change the function call to
printf("%d",abc(1));
The output is some garbage value.
How does parameter passing work here? Please explain.
The unexpected results you are getting are not resulting from the "parameter passing", but from the fact that the abc function does not return any value. You should use return k; statement to get the output you are expecting. But as for parameter passing, they are passed by value, i.e. the passed value is copied to a temporary location k (visible in the function only), and not modified outside of it.
The code example you have passes a by value. You could think of it as a copy of a. You code modified with comments:
#include<stdio.h>
int abc(int k)
{
// k is a copy of a, it is not a, since k is a copy, it has the
// value of a at the point of the copy. So, k is 1
k++; // k is now 2
return k; // return the computed value to the caller and destroy k
}
int main()
{
int a=1;
// as previously written, without the return statement in abc()
// this function returned nothing. So, the compiler just arranges
// for something to be used from the stack where the return would
// have placed 2. (I'm not terribly familiar
// with assembly and so I'm not sure which register it would use).
// That's why you get non-nonsensical data, whatever is in memory is
// what you get and without the return statement, there's nothing
// meaningful there.
// Also, as I commented above, abc() takes a **copy** of a. Thus,
// the contents of a are unmodified. See how the printf() is
// changed. What does it print?
printf("%d %d",abc(a), a);
return 0;
}
I'm trying to use a separate function to input data using scanf() (outside of main). This new function is supposed to print a line and then receive input from the user. However something appears to be going awry between the scanf in the function and the printf() function in the main that I am testing it with.
I believe that I am receiving a pointer from the function but certain compiler warning are making me wonder if my assumption about the pointer is even correct.
I am confused by the output of this code:
#include <stdio.h>
void set_Info(void);
int main()
{
int scanNum = 0;
set_Info();
printf("%d", &scanNum);
return 0;
}
void set_Info(void) /* start of function definition */
{
int scanNum;
printf("Scan test, enter a number");
scanf("%d",&scanNum);
}
If I provide a number, say 2, the result of the printf statement in the main() is:
2665560
Now, in so far as I am able to tell that output appears to me like a memory address so what i attempted to do to fix that is dereference the pointer in main like so :
int scanNum = 0;
int scanNumHolder;
set_Info();
scanNumHolder = *scanNum;
printf("%d", &scanNumHolder);
I believe that this code makes scanNum variable to become assigned to the dereferenced value of scanNum. However I get the same output as above when I do this. Which leads me to believe one of two things. Either that I am not correctly dereferencing scanNum, or that scanNum is not in fact a pointer at all in this situation.
The most common error I receive from the compiler is:
error: invalid type argument of unary ‘*’ (have ‘int’)
Which makes sense, I suppose, if I'm attempting to treat an int value as a pointer.
If it is the case that scanNum is not being dereferenced correctly, how can I achieve this?
Thank you for the help
*Update
Thanks for the help.
Just to recap
My set_info function needs to be passed an address parameter. The reason an address parameter has to be used is because the local memory of a function is erased after the function call ends. So in order to do work a variable declared in the main function, I pass the address of the variable in question so that when the function ends the changes are not lost.
Inside the main function, when set_info is called with &scanNum as the argument, it passes a reference tp the variable so that it can be assigned the value generated by the scanf statement in the function.
I realize that what I was doing wrong as correctly pointed out by the awesome people of SO, is that I am trying to call set_info like it returns a value but in fact changes the variable like I actually want.
Thanks again for the help!
This function:
void set_Info(void)
{
int scanNum;
scanf("%d", &scanNum);
}
reads the integral number from the standard input and stores it into scanNum variable, which is local variable with automatic storage duration that exists only within the scope of this function.
And the body of your main:
int scanNum = 0;
set_Info();
printf("%d", &scanNum);
defines a local variable called scanNum, then calls a set_Info() function which doesn't affect scanNum defined in main in any way and then it prints the address of scanNum variable.
This is what you are trying to do:
void set_Info(int* num)
{
// read an integer and store it into int that num points to:
scanf("%d", num);
}
int main()
{
int scanNum = 0;
// pass the address of scanNum to set_Info function so that
// changes to scanNum are visible in the body of main as well:
set_Info(&scanNum);
printf("%d", scanNum);
return 0;
}
I also recommend you spend more time reading some book with C basics before you'll continue programming :)
I would pass in the variable into your set_Info function, so that it knows where to save the data. This would then allow you to scan multiple values, and you would simple increment the pointer. Be sure to pass the variable address into set_Info() using &variableName, since that function expects a pointer
#include <stdio.h>
void set_Info(int *pScanNum);
int main()
{
int scanNum = 0;
set_Info(&scanNum);
printf("%d", scanNum);
return 0;
}
//Pass in the pointer to scanNum
void set_Info(int *pScanNum)
{
printf("Scan test, enter a number");
scanf("%d",pScanNum);
}
Get rid of your ampersand! Printf wants an integer not a pointer.
printf("%d", scanNum);
And as liho said, you need to return scanNum from set_info so you can get at it outside of the function.
int scanNum = set_Info();