C での 0 または 1 による変数の暗黙的な初期化
C プログラミング言語では、値を割り当てる前に変数を宣言する必要があります。 例えば：
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
// 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 = %d nn ' a b ); printf ( 'arr[0] = %d n arr[1] = %d n arr[2] = %d' ' nn ' arr [ 0 ] arr [ 1 ] arr [ 2 ]); printf ( 'i = %d n ' i ); return 0 ; }
出力： 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] = %d n ' i arr [ i ]); } return 0 ; }
出力： arr[0] = 1 arr[1] = 2 arr[2] = 3 arr[3] = 0 arr[4] = 0クイズの作成
