As i want to find array size dynamically in function, i used sizeof operator. But i got some unexpected result.
#include<stdio.h> double i; int main() { (int)(float)(char)开发者_开发技巧 i; printf(\"%d\", sizeof((int)(float)(char)i));
I\'ve declared this variable: float (**explosions)[4]; This will point to a memory block of pointers to memory blocks for float arrays with 4 floats.
What are the size guarantees开发者_如何学JAVA on int in Single UNIX or POSIX? This sure is a FAQ, but I can\'t find the answer...With icecrime\'s answer, and a bit further searching on my side, I got
class CHaraICICCC { int i; char c1; int j; char c2; char c3; char c4; }; class CHaraIICCCC { int i; int j; char c1;
Ok while sizeof(Myenum) and sizeof(int) works, I would like to use sizeof(object), but I don\'t want the size of the object, but the size of pointer... only for portability reason, I need to know if i
Consider the following code #include <stdio.h> void print(char string[]){ printf(\"%s:%d\\n\",string,sizeof(string));
My application has a defined structure: typedef struct zsSysVersionMsg_tag { WORD cmd; BYTE len; } zsSysVersionMsg_t;
I suppose much like standard arrays (i.e integer arrays), when you pass an array of structs you must pass the size of the array with it. However what I do not yet understand is that when you take size
I happened to ask myself a question about arrays in c++. Well, we all know that arrays are fixed collections of something, I say fixed because it is necessary to declare array length when defining arr
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