Good day
I'm trying to write a swap function code but it's not working
Any help will be highly appreciated!!Code:int swapfunction (int a, int b) { int i; int temp; a=100;b=50; temp=a; a=b; b=temp; int c[2]={a,b}; return(c[i]); } int main () { int num1,i, num2; int results[2]; results[2]=swapfunction(num1,num2); num1=12; num2=13; for(i=0;i<=1;++i) printf("%d\t",results[i]); return(0); }
Best
Ama
You can't return arrays by value.
The normal approach to swapping would be to pass two pointers to the things you want swapped.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
Thank you for your interest
I'm new to C and a more detailed answer would be very welcomed!!!
This does not change the value in main.
This changes a in main, but you can only return one thing (integral types, float types, pointers, structs, unions) at a time.Code:void foo ( int a ) { a = 42; } int main ( ) { int a = 0; foo(a); // a is still 0 }
This also changes a in main, but using a pointer.Code:int foo ( ) { return 42; } int main ( ) { int a = 0; a = foo(); // a is now 42 }
Knowing that you can have as many parameters as you like(*), consider what this might meanCode:void foo ( int *a ) { *a = 42; } int main ( ) { int a = 0; foo(&a); // a is now 42 }
void swap ( int *a, int *b )
(*) the C99 standard states that a compiler should support at least 127 parameters.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
