sizeof

问题 This question already has answers here : '%d' expects argument of type 'int', but argument 2 has type 'long unsigned int' [-Wformat=] [duplicate] (3 answers) Closed 7 years ago . I am getting a warning warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘long unsigned int’ [-Wformat] I am writing a very basic program which gives the size of the data type but in the Linux environment I am getting this warning whereas in Visual Studio, the program works without any

问题 This program: struct alignas(4) foo {}; int main() { return sizeof(foo); } returns 4, with GCC 10.1 and clang 10.1, and icc 19.0.1 . That makes me wonder - is it mandatory for alignas() to affect sizeof() this way? i.e. increase the size beyond what the structure would originally be sized at? Or - is this change just the implementation's prerogative? 回答1: is it mandatory for alignas() to affect sizeof() this way? i.e. increase the size beyond what the structure would originally be sized at?

问题 This program: struct alignas(4) foo {}; int main() { return sizeof(foo); } returns 4, with GCC 10.1 and clang 10.1, and icc 19.0.1 . That makes me wonder - is it mandatory for alignas() to affect sizeof() this way? i.e. increase the size beyond what the structure would originally be sized at? Or - is this change just the implementation's prerogative? 回答1: is it mandatory for alignas() to affect sizeof() this way? i.e. increase the size beyond what the structure would originally be sized at?

问题 The ISO C standard says that: sizeof(char) <= sizeof(short) <= sizeof(int) <= sizeof(long) I am using GCC-8 on BIT Linux mint (19.1) and the size of long int is 8 . I am using an app which uses GCC 7 and the compiler is 64-bit. The size of long int is 4 . Does the compiler or the operating system define the size of a long int ? 回答1: The compiler calls all the shots. The operating system just runs the resulting binary. That being said, the compiler will normally make an executable the

问题 I tried the following program struct temp{ int ab; int cd; }; int main(int argc, char **argv) { struct temp *ptr1; printf("Sizeof(struct temp)= %d\n", sizeof(struct temp)); printf("Sizeof(*ptr1)= %d\n", sizeof(*ptr1)); return 0; } Output Sizeof(struct temp)= 8 Sizeof(*ptr1)= 8 In Linux I have observed at many places the usage of sizeof(*pointer) . If both are same why is sizeof(*pointer) used ?? 回答1: I generally prefer sizeof(*ptr) because if you later go back and change your code such that

问题 Is sizeof(Type) always divisible by alignof(Type) such that this statement will always be true? sizeof(Type) % alignof(Type) == 0 回答1: Yes, sizeof(Type) % alignof(Type) == 0 is true for all class types. The standard draft says: [dcl.array] ... An object of array type contains a contiguously allocated non-empty set of N subobjects of type T. [expr.sizeof] ... When applied to a class, the result is the number of bytes in an object of that class including any padding required for placing objects

问题 Is sizeof(Type) always divisible by alignof(Type) such that this statement will always be true? sizeof(Type) % alignof(Type) == 0 回答1: Yes, sizeof(Type) % alignof(Type) == 0 is true for all class types. The standard draft says: [dcl.array] ... An object of array type contains a contiguously allocated non-empty set of N subobjects of type T. [expr.sizeof] ... When applied to a class, the result is the number of bytes in an object of that class including any padding required for placing objects

问题 suppose i have a function like this int writetofile(wstring name, any sdata){ ... return error; } This function have no idea about what data would be stored but would need to know the size of the data stored in sdata . Although it is easy to determine the type of data stored in the sdata but i don't think there is some easy way to know about the size of data in sdata . i have a data structure which has members of type wstring . Now We cant write that data structure directly to the file as it

问题 This question already has answers here : malloc in C: same sizeof before and after? (5 answers) Determine size of dynamically allocated memory in C (15 answers) Closed 6 days ago . In this program I have located char pointer. I have located 1 byte of memory in heap. But when I print its size it shows 4 bytes. please help. //CODES// #include <stdio.h> #include <stdlib.h> int main(){ char *ptr; ptr=(char*)malloc(1); printf("The size is %d.",sizeof(ptr)); } 来源： https://stackoverflow.com

来源： https://stackoverflow.com/questions/54054427/different-between-c-struct-bitfields-on-char-and-on-int