Operator & Expression

Theory

Execute this line and think, what is happening...

#include <stdio.h>
int main(void) {
  int a, b, c, d, e;
  e = 7;
  printf("e = %d\n", e);
  c = d = e;
  printf("c = %d, d = %d, e =  %d\n", c, d, e);
  a = b = c = d = e = -3; // last "value" or "value of variable" (-3) is going to be assigned to all variables before the last one
  printf("%d %d %d %d %d ", a, b, c, d, e);
}

similarly,

#include <stdio.h>
int main(void) {
  int a = 1, b = 2, c = -3, d = 0, e = b, f = e;
  printf("%d %d %d %d %d %d", a, b, c, d, e, f);
}

You can assign multiple statements in one line with comma. Luckily, you wont have to write multiple lines with semicolon.

#include <stdio.h>
int main(void) {
  printf("Hello "), printf("World"), putchar('\n'), printf("input any int value:");
  int a = 0;
  scanf("%d", &a), printf("\na is %d...", a);
}

this trick is helpful for situation like:

// Note that, this is a cuda code, do not get confused it with a C Programming language code
// the purpose of presenting it is to let you know that you can use this trick in scenarios like this...

#include <cstdio>
#include <cstdlib>
#include <cuda.h>

#define SIZE 1024

__global__ void VectorAdd(int *a, int *b, int *c, int n) {
  int i = threadIdx.x;
  if (i < n) {
    c[i] = a[i] + b[i];
  }
}

/*
".cu": "cd $dir && nvcc $fileName -o $fileNameWithoutExt -arch=sm_86 &&
$dir$fileNameWithoutExt",
*/
int main(void) {
  int *a, *b, *c;
   
  cudaMallocManaged(&a, SIZE * sizeof(int)), cudaMallocManaged(&b, SIZE * sizeof(int)), cudaMallocManaged(&c, SIZE * sizeof(int));

  for (int i = 0; i < SIZE; i++) {
    a[i] = b[i] = i;
    c[i] = 0;
  }


  int threadsPerBlock = 256;
  int blocksPerGrid = (SIZE + threadsPerBlock - 1) / threadsPerBlock;

  // VectorAdd<<<1, SIZE>>>(a, b, c, SIZE); // <<<1, SIZE>>> WAS AN ERROR
  VectorAdd<<<blocksPerGrid, threadsPerBlock>>>(a, b, c, SIZE);

  // Kernel execution happens here

  cudaDeviceSynchronize(); // Wait for kernel to finish

  for (int i = 0; i < 10; i++) {
    printf("c[%d] = %d\n", i, c[i]);
  }

  cudaFree(a), cudaFree(b), cudaFree(c);

} // WORKING

Problems

• Click on the title of the problem to submit your solution before looking at our solution.
• We discourage looking at our solution before attempting to solve it by yourself first.

1. Simple Calculator ↗️

   Given two numbers X and Y. Print the summation and multiplication and subtraction of these 2 numbers.

   Input

   Only one line containing two separated numbers \(X, Y (1  ≤  X, Y  ≤  10 ^ 5)\)

   Output

   Print 3 lines that contain the following in the same order:

   1. "\(X + Y =\) summation result" without quotes.
   2. "\(X * Y =\) multiplication result" without quotes.
   3. "\(X - Y =\) subtraction result" without quotes.

   Examples

   Input

   5 10
   

   Output

   5 + 10 = 15
   5 * 10 = 50
   5 - 10 = -5
   

   Notes

   Be careful with spaces.

   Hint:

   Press the eye icon to reveal hint.
    Use long long int instead of int
    Don't forget to use the designated format specifier for long long int (%lld)
   

   Solution

   // Press the eye icon to reveal solution.
   #include <stdio.h>
   
   int main()
   {
       long long x, y;
   
       scanf("%lld %lld", &x, &y);
   
       printf("%lld + %lld = %lld\n", x, y, x + y);
       printf("%lld * %lld = %lld\n", x, y, x * y);
       printf("%lld - %lld = %lld\n", x, y, x - y);
   
       return 0;
   }
   

2. Difference ↗️

   Hint:

   Press the eye icon to reveal hint.
    Use long long int instead of int
    Don't forget to use the designated format specifier for long long int (%lld)
   

   Solution

   // Press the eye icon to reveal solution.
   #include <stdio.h>
   
   int main()
   {
       long long a, b, c, d, x;
   
       scanf("%lld %lld %lld %lld", &a, &b, &c, &d);
   
       x = (a * b) - (c * d);
   
       printf("Difference = %lld\n", x);
   
       return 0;
   }