# C programming

Discussion in 'C' started by ROBAODOM, Nov 26, 2011.

1. ### ROBAODOMNew Member

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Write a C program to output the following multiplication table of a number,when the inputs are keyed in from the keyboard.

5x1=5
5x2=10
5x3=15
-------
-------
5x10=50

2. ### pravinkandalaNew Member

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Code:
```#include<conio.h>
#include<stdio.h>
main()
{
float a,b,c;
clrscr();
printf("enter a and b values\n");
scanf("%f%f",&a,&b);
c=a*b;
printf("Multiplication of a and b is:%f",c);
getch();
}```

3. ### pravinkandalaNew Member

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Output:

enter a and b values
5
1
Multiplication of a and b is: 5

//i.e., the number and multiplier , we asume it as a and b

4. ### newoneNew Member

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#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a,b;
printf("enter value");
scanf("%d", &a);
printf("enter value again");
scanf("%d", &b);
printf("%d + %d = %d\n", a, b, a*b);
getch();
}

output
enter value
5
enter another value
1
5*1=5

5. ### newoneNew Member

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For multiplication table of a number,when the inputs are keyed in from the keyboard:

#include<stdio.h>
#include<conio.h>
void main()
{
clrscr();
int a,b;
printf("enter the values");
scanf("%d", &a);
scanf("%2d", &b);
while(b<=10)
{
printf("%d * %2d = %2d\n", a, b, a*b);
b=b+1;
}
getch();
}

output
enter the values
5
1
5*1=5
5*2=10
5*3=15
5*4=20
5*5=25
...........
...........
5*10=50

6. ### hobbyistNew Member

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Code:
```#include <stdio.h>
#include <string.h>
#include <limits.h>

int main(void) {

char buf[10] = { 0 };
int i, num;

do {

/* wait for valid numerical input */

printf("Enter table multiplier: ");
fgets(buf, sizeof buf, stdin);

} while((sscanf(buf, "%d", &num)) != 1);

for(i=1; (i * num) < INT_MAX; ++i) {

if(i % 10 == 0) {
puts("\npress enter");
getchar(); // pause output
}

printf("%3d %2c %3d = %3d\n", num, 'X', i, i * num);
}

return 0;
}```
:P