C++11, §5.3.3 ¶6 The result of sizeof and sizeof… is a constant of type std::size_t. [ Note: std::size_t is defined in the standard header (18.2). — end note ] You can also do a quick check: which correctly outputs 1 on my machine. As @Adam D. Ruppe said in the comment, probably the compiler does not complain because, since it already knows the result, … Read more
Yes, it’s a matter of style, because you’d expect sizeof(char) to always be one. On the other hand, it’s very much an idiom to use sizeof(foo) when doing a malloc, and most importantly it makes the code self documenting. Also better for maintenance, perhaps. If you were switching from char to wchar, you’d switch to … Read more
A long, as far as I know, is 8 bytes. Is this correct? No, this is not correct. The C and C++ specifications only state that long must be greater than or equal to 32 bits. int can be smaller, but on many platforms, in C and C++, long and int are both 32 bits. This is a very good reason … Read more
Given an arbitrary type (I’ve chosen char here, but that is for sake of concrete example): You can use either of these expressions: Leading to a malloc() call such as: The last version has some benefits in that if the type of ppc changes, the expression still allocates the correct space.
char is a character and sizeof(char) is defined to be 1. (N1570 6.5.3.4 The sizeof and _Alignof operators, paragraph 4) char* is a pointer to a character and sizeof(char*) depends on the environment. It is typically 4 in 32-bit environment and 8 in 64-bit environment. In typical environment where sizeof(char*) > sizeof(char), malloc(sizeof(char*)*len + 1) will (at least try to) allocate more memory than malloc(sizeof(char)*len + 1) if len is small … Read more
How do I determine the size of an object in Python? The answer, “Just use sys.getsizeof“, is not a complete answer. That answer does work for builtin objects directly, but it does not account for what those objects may contain, specifically, what types, such as custom objects, tuples, lists, dicts, and sets contain. They can contain instances each … Read more
he guarantee you get is that sizeof(char) == 1. There are no other guarantees, including no guarantee that sizeof(int *) == sizeof(double *). In practice, pointers will be size 2 on a 16-bit system (if you can find one), 4 on a 32-bit system, and 8 on a 64-bit system, but there’s nothing to be gained in … Read more
Answer: sizeof returns the size of the type in bytes. Example: sizeof(char) is 100% guaranteed to be 1, but this does not mean, that it’s one octet (8 bits). Proved by the standard: in 6.5.3.4, point 2: The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The … Read more
Because 3.0 is a double. See C syntax Floating point types. Floating-point constants may be written in decimal notation, e.g. 1.23. Scientific notation may be used by adding e or E followed by a decimal exponent, e.g. 1.23e2 (which has the value 123). Either a decimal point or an exponent is required (otherwise, the number is … Read more
If you have an array then sizeof(array) returns the number of bytes the array occupies. Since each element can take more than 1 byte of space, you have to divide the result with the size of one element (sizeof(array[0])). This gives you number of elements in the array. Example: LIVE EXAMPLE Note that if you … Read more