swap (int *a, int *b) { *a ^= *b; *b ^= *a; *a ^= *b; }
Try the triangle program on a search engine. Replace numbers with stars and that should do the trick
`<?php` then a `?>` and also `<?=` and `?>` are the only compliant methods now that PHP 7 is out.
You can use: require('filename.extension'); Or: include('filename.extension');
You are trying to compare two things that are completely different. PHP is a programming language, and QA testing is a step in the creation of products.
swap (int *a, int *b) { *a ^= *b; *b ^= *a; *a ^= *b; }
You cannot swap two numbers using call by value, because the called function does not have access to the original copy of the numbers.Swap with call by reference... This routine uses exclusive or swap without temporary variable.void swap (int *a, int *b) {*a ^= *b;*b ^= *a;*a ^= *b;return;}
By using the algorithm of bitwise EORing (Exclusive ORing) the numbers together:If the two numbers are X and Y, then to swap them:X = X EOR YY = Y EOR XX =X EOR Ywill swap them.With knowledge of that algorithm, one then uses the syntax of Javascript to implement it.
//lets assume a = 10; b = 20; a = a^b; b = a^b; a = a^b;
To swap two numbers N1 and N2, using a third variable T... T = N1; N1 = N2; N2 = T;
You have to pass the address of the variables.void swap (int *pa, int *pb){...}
void swap (int *a, int *b) { *a ^= *b; *b ^= *a; *a ^= *b; return; }
The using of term 'call-by-reference' implies function-call, so please rethink your question...
void swap(int& a, int& b ) { a^=b^=a^=b; }
To swap two variables without using a third variable, use exclusive or manipulation... a ^= b; b ^= a; a ^= b;
The required c program is given below /*Swapping(interchange) the two entered numbers*/ #include<stdio.h> main() { /*Without using third variable*/ int a,b,t; printf("Enter a:"); scanf("%d",&a); printf("Enter b:"); scanf("%d",&b); a=a+b; b=a-b; a=a-b; printf("\n After swapping without using third variable"); printf("\na=%d\nb=%d",a,b); }
a += b; b -= a; a -= b;