This is all 100% standardized. C17 6.10.8.1: __DATE__ The date of translation of the preprocessing translation unit: a character string literal of the form “Mmm dd yyyy” … and the first character of dd is a space character if the value is less than 10. … __TIME__ The time of translation of the preprocessing translation … Read more
In C, the type of a character constant like ‘a’ is actually an int, with size of 4 (or some other implementation-dependent value). In C++, the type is char, with size of 1. This is one of many small differences between the two languages.
C standard forbids it – when compiled with gcc -pedantic, it produces invalid application of ‘sizeof’ to a function type warning. However gcc compiles it and returns 1 for sizeof(main), and it is not a size of function pointer. It seems to be compiler-dependent.
sizeof(), while looking like a function call, is actually an operator and part of the language core. No include needed. size_t is defined in various headers: stddef.h, string.h, stdlib.h, and stdio.h. Including any one of them is enough to use size_t in your code.
#include <stdio.h> typedef struct { char* c; char b; } a; int main() { printf(“sizeof(a) == %d”, sizeof(a)); } I get “sizeof(a) == 8”, on a 32-bit machine. The total size of the structure will depend on the packing: In my case, the default packing is 4, so ‘c’ takes 4 bytes, ‘b’ takes one … Read more
The actual issue here is that use of unary minus operator, just like the rest of built-in arithmetic operators, is a subject to integral promotions. So surprisingly the result of applying unary minus to size_t will be still size_t and there is no need to blame auto. Counter-example. In this case due to integral promotions … Read more
Yes, size_t n = sizeof(tbl) / sizeof(tbl[0]) is the most typical way to do this. Please note that using int for array sizes is not the best idea.
In C++, the expressions inside sizeof is not evaluated, except for C99’s VLA as mentioned in comments, since this was earlier tagged for C too The sizeof operator is calculated at compile time. Only the type of the expression (that is calculated at compile time) is then used by sizeof. From C++ Standard § 5.3.3 … Read more
Let’s say I have an array arr. When would the following not give the number of elements of the array: sizeof(arr) / sizeof(arr[0])? One thing I’ve often seen new programmers doing this: void f(Sample *arr) { int count = sizeof(arr)/sizeof(arr[0]); //what would be count? 10? } Sample arr[10]; f(arr); So new programmers think the value … Read more
%d prints an integer: it will print the ascii representation of your character. What you need is %c: printf(“%c”, ch); printf(“%d”, ‘\0’); prints the ascii representation of ‘\0’, which is 0 (by escaping 0 you tell the compiler to use the ascii value 0. printf(“%d”, sizeof(‘\n’)); prints 4 because a character literal is an int, … Read more