no obviously not its an exam
A)Int B)Long int C)Unsigned Int D)Float
555
http://www.post.japanpost.jp/int/question/45_en.html
You're dirty minded!
int n1; int n2; int n3; int n4; int n5; int n6; int n7; int n8; int n9; int n10; int n11; int n12; int n13; int n14; int n15; int n16; int n17; int n18; int n19; int n20; int n21; int n22; int n23; int n24; int n25; int n26; int n27; int n28; int n29; int n30;
The phrase 'fortitud int deo' translates from Latin as 'between the strength of God'
// declare a function int* function(int, int); or int* (function)(int, int); // declare a pointer to a function int* (*pointer_to_function)(int, int);
yes. we can sat . because the same method may be used by many banks but implementations may be different. Eg: withDraw(int i) { int maxWithDraw = 10000; if(i>maxWithDraw) { Sop("Not allowed"); } } Eg: withDraw(int i){ ** int maxwithDraw = 20000; ....... } Here we can see that the methods are the same but implementation is different. This is nothing but overriding.
For their names in the English language the answer is 4/12.
printf ("sizeof (int) = %d\n", (int)sizeof (int));
int LCM3 (int a, int b, int c) { return LCM2 (a, LCM2 (b, c)); } int LCM2 (int a, int b) { return a*b/GCD2(a, b); }
int sum(int list[], int arraySize) { int sum=0; for(int i=0; i<arraySize; ++i ) sum+=list[i]; return(sum); }