用户输入后程序结束

Question

Program that I wrote stops after user input, and I don't have any idea why this is happening. User should give 6 integers, and after that program should do some math with it which isn't the case right now, because after inserting integers it ends, with no error.

 #include <stdlib.h>
 #include <stdio.h>


 int main(int argc, char *argv[]) {

    int T,i;
    int x[5];

     scanf("%d", &T);
     while(T > 0) {
         for(i = 0; i <= 4; i++) {
             scanf("%d", &x[i]);
          }
          int size_tax = x[0] + x[1] + x[2];
          int tax[size_tax];

          for(i = 0; i <= size_tax; i++) {
              if(i == 0) {
                   tax[i] = x[0];
                   if(x[4] == 1) {
                       printf("%d", tax[0]);
                       break;
                    } 
              } else if(i > 0 && i <= x[1] - 1) {
                  tax[i] = tax[i-1] + 1;
                  if(x[4] == i + 1) {
                      printf("%d", tax[i]);
                      break;
                   }
              } else if(i > x[1] - 1 && i <= x[2] - 1) {
                  tax[i] = tax[i-1] * 2;
                      if(x[4] == i + 1) {
                      printf("%d", tax[i]);
                      break;
                    }
              }
          }

         if(x[4] > x[0] + x[1] + x[2]) {
              int result = x[3] - 1;
              for(i = x[3]; i <= x[4] - 1; i++) {
                  result = result * x[i];

              }

              printf("%d", result);
        }

         T--;
      }
 } 

Answer 1

Here is the problem:

int tax[size_tax];

for(i = 0; i <= size_tax; i++) {
     ...
}

The declared size of tax is size_tax , so the loop must go up to size_tax-1 , so it should be

for(i = 0; i < size_tax; i++) { // Less than, not less than or equal to
     ...
}

Answer 2

If the input your testing is 1 2 3 4 5 6 then that's perfectly normal. Just insert an else after the last else if. It seems this input won't verify any other condition.