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A int pow(int a,int b)
{
int temp=0,i;
while(b!=0)
{
for(i=0;i<a;i++)
temp=temp+a;
a=temp;
temp=0;
b--;
}
return a;

Warning: Illegal string offset 'name' in /home/prepdo6/gpl4you/discussquessub.php on line 681
A Using recursion, we can do it.

#include

/* assumption: b is greater than or equal to 0*/
int pow(int a, int b)
{
if(b == 0)
return 1;
return multiply(a, pow(a, b-1));
}

/* assumption: y is greater than or equal to 0*/
int multiply(int x, int y)
{
if(y == 0)
return 0;
return (x + multiply(x, y-1));
}

int main()
{
printf("\n %d", pow(5, 3));
getchar();
return 0;
}

Warning: Illegal string offset 'name' in /home/prepdo6/gpl4you/discussquessub.php on line 681
A void main()
{
int a,b,res,t;

cout<<"enter a and b"<<endl;
cin>>a>>b;

res=a;

for(int i=1;i<b;i++)
{
t=res;
for(int j=1;j<a;j++)
{
res+=t;
}
}
cout<<res;
}

Warning: Illegal string offset 'name' in /home/prepdo6/gpl4you/discussquessub.php on line 681
A I think in the given solution no of additions in the func. multiply for higher values of 'b' would be enormous.
Therefore I propose a slightly different solution(90% same though).

#include<stdio.h>
#include<conio.h>

int add(int x, int y)
{
if(x==1)
return y;
return (y + add(x-1,y));
}

int power(int a,int b)
{
if(b==0)
return 1;
if(b==1)
return a;
return add(a, power(a,b-1));
}

int main()
{
int a, b;

printf("\nEnter the two numbers :\n");
scanf("%d%d",&a,&b);

printf("pow(a,b) = %d",power(a,b));

getchar();
return 0;

}