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Niejawna inicjalizacja zmiennych za pomocą 0 lub 1 w C

W języku programowania C zmienne należy zadeklarować przed przypisaniem im wartości. Na przykład:
 // declaration of variable a and // initializing it with 0. int a = 0; // declaring array arr and initializing // all the values of arr as 0. int arr[5] = {0}; 
However variables can be assigned with 0 or 1 without even declaring them. Let us see an example to see how it can be done: C
#include  #include  // implicit initialization of variables a b arr[3]; // value of i is initialized to 1 int main(i) {  printf('a = %d b = %dnn' a b);  printf('arr[0] = %d narr[1] = %d narr[2] = %d'  'nn' arr[0] arr[1] arr[2]);  printf('i = %dn' i);  return 0; } 
Wyjście:
a = 0 b = 0 arr[0] = 0 arr[1] = 0 arr[2] = 0 i = 1 

In an array if fewer elements are used than the specified size of the array then the remaining elements will be set by default to 0. Let us see another example to illustrate this. C
#include  #include  int main() {  // size of the array is 5 but only array[0]  // array[1] and array[2] are initialized  int arr[5] = { 1 2 3 };  // printing all the elements of the array  int i;  for (i = 0; i < 5; i++) {  printf('arr[%d] = %dn' i arr[i]);  }  return 0; } 
Wyjście:
arr[0] = 1 arr[1] = 2 arr[2] = 3 arr[3] = 0 arr[4] = 0 
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