#swapping variables

32 messages · Page 1 of 1 (latest)

pastel coyote
# 
#include <iostream>
using namespace std;



void swap(int x; int y) {
    
    int temp;
    x = y;
    y = temp;
    
}


int main() {


   int x = 1;
   int y = 2;
   
   swap(x,y);

   cout<<x<<y<<endl;

    return 0;
}```
errant whaleBOT
#

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pastel coyote
#

It is giving errors

honest pebble
#

what is the expected behaviour? your function does not modify any variables passed to it by the caller

pastel coyote
#

Can you explain?

# 
main.cpp:6:16: error: expected ‘)’ before ‘;’ token 6 | void swap(int x; int y) { | ~ ^ | ) main.cpp:6:23: error: expected initializer before ‘)’ token 6 | void swap(int x; int y) { | ^ ```


The error
tame grove
#

;

honest pebble
#

you aren’t supposed to have that semicolon there

#

in function declarations parameters are comma-seperated

pastel coyote
#

Oh

#

I'm dumb.

honest pebble
#

is this your first time writing a program?

#

there are other issues

tame grove
#

so shouldn't be first time

honest pebble
#

your function only modifies its own local copies of its arguments, not the versions you see printing out

honest pebble
pastel coyote
honest pebble
#

well you have to pass the variable by pointer or reference

#

do you know what those are

pastel coyote
#

Yeah

honest pebble
#

you are currently using the 2nd behaviour, you want the first one

pastel coyote
# 
void swap(int *x, int *y){
   int temp;
   x = temp;
   x = y;
   y = temp;
}```

Something like that?
#

Ahh , pointers are hard

honest pebble
#

int* temp

pastel coyote
#

!solved

errant whaleBOT
#

Thank you and let us know if you have any more questions!

#

[SOLVED] swapping variables

honest pebble
#

now try to do it without the temp

pastel coyote
# 
int temp = x*;
x* = y*;
y* = temp;```
pastel coyote