Best Answer

main()

{

in a,b,c,d;

printf("\n enter any four numbers");

scanf("%d%d%d%d",&a,&b,&c,&d);

if(a==ba==cb==cb==dc==d)

printf("\n numbers must be different");

else

{

if(a>b&&a>c&&a>b)

X=b>c&&b>d?b;c>d?c:d;

else if(b>c&&b>d)

x=a>c&&a>b?a:c<d?c:d;

else if(c>d)

x=a>b&&a>d?a:b>d?b:d;

else

x=a>b&&a>c?a:b>c?b:c;

printf("\n %d is second max vaalue",x);

{

}

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I will just write the function - the header stuff is easy.

int largest (int numbers[], int n)

{

int i;

int biggest=numbers[0]; /* First assign the first number to the biggest */

for (i=1; i<n; i++) /* we loop through from the 2nd element to the last */

{

if (numbers[i] > biggest) /* if we find one larger then we make it biggest */

biggest = numbers[i];

}

return biggest; /* Note: this may or may not work, you see biggest is defined ONLY for the function if you try to send it outside of the function it may not work (depending on the compiler). I put this here because it helps explain the program BEST! However, you may need to use a static variable or a global variable. */

}

#include<stdio.h>

#include<conio.h>

void main()

{

int i, n, a, b, num;

a = b = 0;

clrscr();

printf("Enter the Maximum amount of Numbers :: ");

scanf("%d", &n);

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

{

printf("\nEnter the Number :: ");

scanf("%d", &num);

if (num > a)

{

b = a;

a = num;

}

else if (num > b)

b = num;

}

printf("\n\n Second Highest Number is :: %d",b);

getch();

}

void main()

{

int i,k;

int a[5];

for(j=0;j<5;j++)

{

a[j]=j;

}

for(i=0;i<1;i++)

{

if(a[i]<a[i+1])

{

k=a[i];

a[i]=a[i+1];

a[i+1]=k;

}

}

printf("%d",a[1]);

}

In order to find the second largest you first need to find the largest. The following program demonstrates one such method using an array of doubles. An alternative method is to sort the numbers in descending order then look for the first value that is not equal to the first number, however sorting will add an unnecessary and expensive overhead.

#include<stdio.h>

#include<float.h>

const double lowest = -DBL_MAX;

double find_largest (double* a, const unsigned size) {

if (size<1) return lowest;

double largest = a[0];

unsigned i = 0;

while (++i<size) {

if (largest<a[i])

largest = a[i];

}

return largest;

}

double find_second_largest (double* a, const unsigned size) {

if (size<2) return lowest;

const double largest = find_largest (a, size);

double second = lowest;

unsigned i = 0;

while (++i<size) {

if (a[i]<largest && second<a[i])

second = a[i];

}

return second;

}

int main (void) {

double a[10];

a[0] = 0.5;

a[1] = 0.7;

a[2] = 0.1;

a[3] = 0.9;

a[4] = 0.6;

a[5] = 0.4;

a[6] = 0.2;

a[7] = 0.8;

a[8] = 0.3;

a[9] = 0.0;

double x = find_second_largest (a, 10);

if (x==lowest) {

/* error! */

}

printf ("%f is the second largest\n", x);

return 0;

}

#include<stdio.h>

#include<conio.h>

main()

{

int a[100];

int i,n,slargest,largest,ssmallest,smallest;

printf("enter the number of elements");

scanf("%d",&n);

printf("\nenter the array elements");

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

scanf("%d",&a[i]);

if(a[0]>a[1])

{

largest=a[0];

slargest=a[1];

}

else

{

largest=a[1];

slargest=a[0];

}

for(i=2;i<n;i++)

if(a[i]>largest)

{

slargest=largest;

largest=a[i];

}

else

if(a[i]>slargest)

slargest=a[i];

if(a[0]<a[1])

{

smallest=a[0];

ssmallest=a[1];

}

else

{

smallest=a[1];

ssmallest=a[0];

}

for(i=2;i<n;i++)

if(a[i]<smallest)

{

ssmallest=smallest;

smallest=a[i];

}

else

if(a[i]<ssmallest)

ssmallest=a[i];

printf("\n second largest element=%d",slargest);

printf("\n second smallest element =%d",ssmallest);

getch();

}

You would use a comparison expression across multiple numbers, keeping track of the largest number and comparing it with each number supplied. To do this with a function, supply the function with an array of numbers as well as the number of elements in an array.

Then loop through the array. Keeping track of which number is largest, compare each element in the array with the largest number. If the element's value is larger than the currently stored number, store that element's number as the current number.

Once the loop is finished, simply return the largest number in the array. If no numbers have been supplied, you can return 0 or -1, assuming they are not expected to be in the array.

Find below a collection of related links containing tutorials and quick references that'll help you out immeasurably with C!

146..4 is the second largest digit from a no

Q: C program to find the second highest number from a set of numbers?

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int x; //first number int y; //second number int z = x*y;

/*PROGRAM TO ACCEPT TWO NUMBERS FROM THE USER AND PRINT THEIR MULTIPLICATION. */ #include<stdio.h> #include<conio.h> void main() { int a, b, c; // Declaration of Variables. Variables 'a' & 'b' to hold first & second number. And 'c' to hold result. clrscr(); // To clear the output screen every time program is executed. printf("\n Enter the first number : "); scanf("%d", &a); // To accept the first number. printf("\n Enter the second number : "); scanf("%d", &b); // To accept the second number. c = a*b; // Logic to get the product of the entered two numbers. printf("\n Multiplication of %d & %d = %d", a, b, c); // Displaying result. getch(); // To hold the output screen. }

Since all decent programming languages have the multiplication operator, you simply multiply them. If you want to go into a bit more detail, it would be a bit like this: Ask user for number "a" Ask user for number "b" Calculate result = a * b Show result (End) Use the appropriate flow chart symbols for input, etc.

// largest = largest of a, b, c public class largest { public static void main(String args[]) { int a,b,c,largest; a=0; b=0; c=0; a=Integer.parseInt(args[0]); b=Integer.parseInt(args[1]); c=Integer.parseInt(args[2]); largest=a>b?(a>c?a:c):(b>c?b:c); System.out.println("The largest no. of "+a+","+b+"and"+c+"is"+largest); } }

The VBScript/Visual BASIC Script programming language uses both letters/numbers to create programs with. So, I'm not exactly sure what you mean by making a VBScript which uses only numbers, alone...? As the 'numpad' is made up of just numbers. You need to explain yourself a little bit better, I'm afraid. Did you mean, use VBScript to create a 'calculator program' with, possibly? Anyway, here is a simple VBScript calculator program... num1=CInt(InputBox("Enter the first number: ", "PROGRAM: Add 2 numbers")) num2=CInt(InputBox("Enter the second number: ", "PROGRAM: Add 2 numbers")) MsgBox("The answer is: " & num1+num2) Type the above code into Windows Notepad; then, save it as... add2num.vbs ...go and find the file you just saved; and, left double click on it to make the program RUN/execute. When the program runs you should see... 1. Enter the 1st number: (you can use the NumPad to type in a number) 2. Enter the 2nd number: (you can use the NumPad to type in a next number) 3. Finally, you should see a Windows standard message box appear with the sum total result of the two numbers which you did previously type in. To do other sums...then, you can modify this simple calculator using... + ...plus sign - ...minus sign * ...multiplication sign / ...division sign

Related questions

While there is no "highest number" as numbers themselves can increase infinitely, the second highest named number is a centillion.

The answer will depend on what the highest and lowest numbers are!The answer will depend on what the highest and lowest numbers are!The answer will depend on what the highest and lowest numbers are!The answer will depend on what the highest and lowest numbers are!

There is no highest number, highest prime number, or highest composite number. Therefore, there can't be a second-highest, third-highest, etc., of any of these types of numbers, either.

Find the highest number, eliminate it from the list, find the highest number of the remaining numbers.Find the highest number, eliminate it from the list, find the highest number of the remaining numbers.Find the highest number, eliminate it from the list, find the highest number of the remaining numbers.Find the highest number, eliminate it from the list, find the highest number of the remaining numbers.

Put the numbers in order from least to greatest. mark the lowest number, then the highest, then the second lowest , then second highest, then move inward crossing off the numbers, until you reach the center number.

Take the first number in the set and compare it against the second. If the second is higher, take that number otherwise keep the first Repeat with all the numbers in the set, comparing the highest found so far with the next number in the set, taking the higher of the two until all the numbers in the set have been checked. The number with which you remain will be the highest.

int x; //first number int y; //second number int z = x*y;

= The sum of two numbers is -42 the first number minus the second number is 52 Find the numbers? =

No.

No

No

If the second number is a multiple of the prime number, than the LCM is the second number. If the second number is not a multiple of the prime number, then the two numbers are relatively prime, and the LCM is the product of the two numbers.