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Types

  • while
    • Entry-controlled loop
  • do...while
    • Exit-controlled loops
  • for

Components

  • Initialization
    • \(n\)
    • \(i\)
  • Loop condition checked
  • Code to execute is run
  • Updation occurs
flowchart LR

i[Initialization] --> c[Condition] --> Code --> Updation --> c
/*
int n = 10;
int i = 1; // comesntenret

for (; ; )
{

  if(i>n)
    break;

  // code

  i++;
}
*/

for (int i = 1, n=10; i<=n; i++)
{

}

int i = 1, n=10;
for (; i<=n; i++)
{
  ;// nothing happens
}

printf("%d", i); 11

for (int i = 1, n=10; i<=n; i++)
  printf("%d\n", i);

1
2
...
9
10

for (int i = 1, n=10; i<=n; i++)
  printf("%d\n", i);
  printf("%d\n", i);

for (int i = 1, n=10; i<=n; i++)
{
  printf("%d\n", i);
}
printf("%d\n", i);

// here

1
2
3
...
9
10
11

for (int i = 1, n=10; i<=n; i++)
{
  printf("%d\n", i);
  printf("%d\n", i);
}

// here

1
1
2
2
...
9
9
10
10

int i, n;
for (i = 1, n=10; i<=n; i++);
  printf("%d\n", i);

int i, n;
for (i = 1, n=10; i<=n; i++)
{
 ; 
}

printf("%d\n", i);

11

int i = 1, n = 10;

// curly brace
// run 10 times
while(i<=n)
{
 print("hi");
 i++;
}

int i = 1, n = 10;
// curly brace

do
{
  // code
  i++;
} while (i<n); // has semi-colon

x=5, y=10

Pyramid example

*
*
*
*

#include <stdio.h>

int main()
{
  int n = 4; // no of lines
  for (int i=1; i<=n; i++) // run the loop 4 times
  {
    printf("*\n");
  }

  return 0;
}

*
**
***
****

#include <stdio.h>

int main()
{
  printf("*\n");
    printf("**\n");
    printf("***\n");
    printf("****\n");

  return 0;
}
  1. DUMB
  2. Not scalable
  3. Not elegant

Hence, we need loops (iterative statements)

#include <iostream.h>

int main()
{
  for (int i=1; i<=n; i++)
  {
    for (int j=1; j<=i; j++)
    {
      printf("*");
    }

    printf("\n");
  }
  return 0;
}

jump statements

  • break
    • exits the current loop
  • continue
    • skips the below code of the current iteration
    • goes to the update segment of (for)loop
    • in other loops, it goes to condition
    • then continues as usual
for (int i=1; i<=5; i++)
{
  if(i==3)
  {
    break;
    printf("%d\n", i);
  }

   printf("%d\n", i);
}

// come here

1
2

for (int i=1; i<=5; i++)
{
  if(i==3)
  {
    continue;
    printf("%d\n", i);
  }

   printf("%d\n", i);
}

1
2
4
5

int i=1;
while(i<=5)
{
  if(i==3)
  {
    continue;
    printf("%d\n", i);
  }

   printf("%d\n", i);

  i++;
}

1
2
(infinite loop)

int i=1;
do
{
  if(i==3)
  {
    continue;
    printf("%d\n", i);
  }

   printf("%d\n", i);

  i++;
} while(i<=5); // has terminator

1
2
(infinite loop)

Loop Printining Questions

int n = 4; // no of lines
int j = 1;
for (int i=1; i<=n; i++) // controls no of lines
{
  printf("%d", j);
  j++;

}

1
2
3
4

int n = 4; // no of lines

for (int i=1; i<=n; i++) // controls no of lines
{
  int j = 1;
  printf("%d", j);
  j++;
}

1
1
1
1
Last Updated: 2023-01-25 ; Contributors: AhmedThahir

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