Display numbers from 1 to 10 using for loop?

Answer 1

the following is C code, should work well enough. If not, you should get the general idea.

int counter;

for (int i=1;i<10;i++)

{

println("Enter a number: ");

cin << counter;

}

println("Total, %i", counter);

Answer 2

#include<stdio.h> void main() { int i,n,f1=0,f2=1;f,sum=0; printf("enmter the n value"); scanf("%d",&n); for(i=1;i<=n;i++) { f1=f2; f2=f; f=f1+f2; sum=sum+f; }//loop end }main end

Answer 3

Place the numbers in an array, then pass the array to a function that sums the squares. Note that you must check for overflow to ensure the returned value is valid.

long long sum_squares (const int a[], const int size, bool* overflow) { int i;

long long sum, product;

*overflow=false;

sum = 0;

for (i=0; i<size; ++i) {

if (*overflow = product_overflow (a[i], a[i])) break;

product = (long long) a[i] * a[i];

if (*overflow = sum_overflow (sum, product)) break;

sum += product;

}

return sum;

}

Example usage:

int main (void) {

bool overflow;

int a[] = {42, 39, 56, INT_MAX }; // a likely overflow

long long sum = sum_squares (a, 4, &overflow)

if (overflow) {

// sum is invalid

} else {

// sum is valid

}

return 0;

}

The overflow test functions have the following definitions:

#include<limits.h> // defines INT_MAX and LLONG_MAX

bool sum_overflow (const long long a, const long long b) {

return a > 0 && b > LLONG_MAX - a;

}

bool product_overflow (const long long a, const long long b) {

return a > 0 && b > LLONG_MAX / a;

}

Answer 4

for(int i = 0; i <= 10; ++i) {

}

Answer 5

int main()

{

for(int i = 1; i <= 10; i++)

std::cout << i << endl;

return 0;

}