CODEWITHSUNDEEP
c
coderx123
coderx123

Reputation: 3

C - if statement is being ignored

h_array[i] is less than 19 for the first few cases. However, it still prints k as 0. Could someone help me out as to why this happens?

  for (int i = 0; i < 10; i++) {
    int j = h_array[i];
    int k = 0;
    if (h_array[i]<19) {
      int k = 20 - j;
    }
    int l = 20;
    while (l>=k) {
      printf ("%d - %d\n\n\n",l,k);
      l--;
    }
  }

Upvotes: 0

Views: 556

Answers (2)

AlphaGoku
AlphaGoku

Reputation: 1030

Its because of the scope visibility of the variable int k

Your code has 2 int k. An outer k and an inner k. When the k inside the if statement goes out of scope, the value being printed is the value of the outer k, which in this case is 0.

The correct solution would be:

k = 20 - j;

Not :

int k = 20 - j;

Upvotes: 1

Arran Cudbard-Bell
Arran Cudbard-Bell

Reputation: 6075

The issue is you've re-declared 'k' inside the body of the if statement. The compiler will usually generate a warning about this.

In this case the 'k' variable with the scope of the condition body is modified, whilst the 'k' in the parent scope retains its original value (0).

Removing the type specifier should fix the issue.

  for (int i = 0; i < 10; i++) {
    int j = h_array[i];
    int k = 0;
    if (h_array[i]<19) {
      k = 20 - j;
    }
    int l = 20;
    while (l>=k) {
      printf ("%d - %d\n\n\n",l,k);
      l--;
    }
  }

Upvotes: 3

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