How do you print 1 to 10 number using for loop?

Answer 1

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

namespace printnumber

{

class Program

{

static void Main(string[] args)

{

int i;

for (i = 0; i < 20; i++)

{

Console.Write(i);

Console.Read();

}

}

}

}

Answer 2

If the numbers are stored in an array called (let's assume) "num(i)" you have something similar to the following, depending on which coding language you use:

for i=1 to 10

print ( num(i) )

next i

Answer 3

In Java:

public void printOddsReverseFrom(int from) {

// If from is even, decrement it so it's odd

if (from % 2 1)

from--;

for (int i = from; i >= 0; i--) {

System.out.println("" + i);

}

}

Answer 4

There are two ways to do this. The naive method is to test each integer in the range to see if it is even:

void print_evens (int min, int max) {

for (; min<=max; ++min) if (!(min%2)) std::cout << min << ' '; }

This is inefficient since, logically, we know that the first number must be odd or even and, if it is odd, we increment it once to make it even. We can then step through the loop in increments of 2 instead of 1. We can cater for this as follows:

void print_evens (int min, int max) { if (min % 2) ++min; // ensure min is even for (; min<=max; min+=2) std::cout << min << ' '; // step through all even numbers

}

Note that the modulo operator (%) returns the remainder after division, thus min % 2 returns 0 if the number is even (no remainder after dividing by 2), or 1 if the number is odd.
There are two ways to do this. The naive method is to test each integer in the range to see if it is even:

void print_evens (int min, int max) {

for (; min<=max; ++min) if (!(min%2)) std::cout << min << ' ';

}

This is inefficient since, logically, we know that the first number must be odd or even and, if it is odd, we increment it once to make it even. We can then step through the loop in increments of 2 instead of 1. We can cater for this as follows:

void print_evens (int min, int max) {

if (min % 2) ++min; // ensure min is even

for (; min<=max; min+=2) std::cout << min << ' '; // step through all even numbers

}

Note that the modulo operator (%) returns the remainder after division, thus min % 2 returns 0 if the number is even (no remainder after dividing by 2), or 1 if the number is odd.

There are two ways to do this. The naive method is to test each integer in the range to see if it is even:

void print_evens (int min, int max) {

for (; min<=max; ++min) if (!(min%2)) std::cout << min << ' ';

}

This is inefficient since, logically, we know that the first number must be odd or even and, if it is odd, we increment it once to make it even. We can then step through the loop in increments of 2 instead of 1. We can cater for this as follows:

void print_evens (int min, int max) {

if (min % 2) ++min; // ensure min is even

for (; min<=max; min+=2) std::cout << min << ' '; // step through all even numbers

}

Note that the modulo operator (%) returns the remainder after division, thus min % 2 returns 0 if the number is even (no remainder after dividing by 2), or 1 if the number is odd.

There are two ways to do this. The naive method is to test each integer in the range to see if it is even:

void print_evens (int min, int max) {

for (; min<=max; ++min) if (!(min%2)) std::cout << min << ' ';

}

This is inefficient since, logically, we know that the first number must be odd or even and, if it is odd, we increment it once to make it even. We can then step through the loop in increments of 2 instead of 1. We can cater for this as follows:

void print_evens (int min, int max) {

if (min % 2) ++min; // ensure min is even

for (; min<=max; min+=2) std::cout << min << ' '; // step through all even numbers

}

Note that the modulo operator (%) returns the remainder after division, thus min % 2 returns 0 if the number is even (no remainder after dividing by 2), or 1 if the number is odd.

Answer 5

There are two ways to do this. The naive method is to test each integer in the range to see if it is even:

void print_evens (int min, int max) {

assert (min<max);

for (; min<=max; ++min) if (!(min%2)) std::cout << min << ' ';

}

This is inefficient since, logically, we know that the first number must be odd or even and, if it is odd, we increment it once to make it even. We can then step through the loop in increments of 2 instead of 1. We can cater for this as follows:

void print_evens (int min, int max) {

assert (min<max);

if (min % 2) ++min; // ensure min is even

for (; min<=max; min+=2) std::cout << min << ' '; // step through all even numbers

}

Note that the modulo operator (%) returns the remainder after division. When we divide by 2 the remainder is either 0 or 1 so min%2 evaluates 1 when min is odd, and 0 when min is even. To reverse the logic, we use the NOT operator (!), such that !0 is true when min is even, and !1 is false when min is odd.

Answer 6

In Java, you might do something like this:

for (int i = 1; i<=10; i++) System.out.println(i);