Calculation of A to the power of B using Functions

Write a function power(a,b), to calculate the value of a raised to b.




#include<stdio.h>
main()
{

int power (a,b);
int a, b, result;

printf("Enter the value of a and b:");
scanf ("%d %d", &a, &b);

result=power(a,b);
printf("%d raised to %d is %d", a, b, result);

}

power (int a, int b)
{

int calculation=1, calc;
for (calc=1; calc <=b; calc++)

{
calculation=calculation*a;
continue;
}

return(calculation);
}





The File can be found at:
Download File

Comments

Niroop said…
may i know whether the CONTINUE statement is required,i think it is not,can u explain me dis plz
Niroop said…
can anybody answer my question plzzz...
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Unknown said…
#include
#include
void roman(void);
main()
{
roman();
getch();
}
void roman()
{
int y,i;
printf("input the year to get its roman numeral:\n");
scanf("%d",&y);
for(i=1;i<=y;i++)
{if(y>=1000)
{
printf("m");
y=y-1000;}
}
for(i=1;i<=y;i++)
{if(y>=500)
{
printf("d");
y=y-500; }
}
for(i=1;i<=y;i++)
{if(y>=100)
{
printf("c");
y=y-100;}
}
for(i=1;i<=y;i++)
{if(y>=50)
{
printf("l");
y=y-50;}
}
for(i=1;i<=y;i++)
{if(y>=10)
{
printf("x");
y=y-10;}
}
for(i=1;i<=y;i++)
{if(y>=5)
{
printf("v");
y=y-5;}
}
for(i=1;i<=y;i++)
{if(y>=1)
{
printf("i");}
}
printf("\n");


}
Unknown said…
#include
#include
void roman(void);
main()
{
roman();
getch();
}
void roman()
{
int y,i;
printf("input the year to get its roman numeral:\n");
scanf("%d",&y);
for(i=1;i<=y;i++)
{if(y>=1000)
{
printf("m");
y=y-1000;}
}
for(i=1;i<=y;i++)
{if(y>=500)
{
printf("d");
y=y-500; }
}
for(i=1;i<=y;i++)
{if(y>=100)
{
printf("c");
y=y-100;}
}
for(i=1;i<=y;i++)
{if(y>=50)
{
printf("l");
y=y-50;}
}
for(i=1;i<=y;i++)
{if(y>=10)
{
printf("x");
y=y-10;}
}
for(i=1;i<=y;i++)
{if(y>=5)
{
printf("v");
y=y-5;}
}
for(i=1;i<=y;i++)
{if(y>=1)
{
printf("i");}
}
printf("\n");


}

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