sizeof

#include <stdio.h> int main() { short int i = 20; char c = 97; printf("%d, %d, %d\n", sizeof(i), sizeof(c), sizeof(c + i)); return 0; } Could some one tell me what happens when sizeof(a+b) "a is short int type & b is char type" Output is : 2, 1, 4 Because of C's standard integral promotion rules , the type of the expression c + i is int , so that's why you're getting the equivalent of sizeof (int) . Note that sizeof is not a function, the parenthesis are only needed when naming a type and to resolve precendence conflicts . Your code coule be written: printf("%zu, %zu, %zu\n", sizeof i, sizeof

I am new to C++ and a found a peculiar feature in C++. I saw the size of an empty is 1 byte, I did some research and found out that is is done because every object must have a distinct address. But I want to know what is the content of that 1 byte that is placed. I know it does not hold the "this" pointer but is it a dummy byte or is there actually some content??? There's no content. It's just a dummy byte. Every class or struct must have its sizeof greater than 0 , ergo your behavior. It's expected and mandated by the standard. It is mandated by the Standard that different objects of the same

sizeof和strlen的区别 strlen计算字符串的长度，以’\0’为字符串结束标志 sizeof是分配的数组实际所占的内存空间大小，不受里面存储内容 wcslen计算宽字符串的长度，以’\0’为字符串结束标志 来源： https://blog.csdn.net/weixin_41481284/article/details/99866559

I'm dealing with some code at work that includes an expression of the form -(sizeof(struct foo)) i.e. the negation of a size_t , and I'm unclear on what the C and C++ standards require of compilers when they see this. Specifically, from looking around here and elsewhere, sizeof returns an unsigned integral value of type size_t . I can't find any clear reference for specified behavior when negating an unsigned integer. Is there any, and if so, what is it? Edit: Ok, so there are some good answers regarding arithmetic on unsigned types, but it's not clear that this is in fact such. When this

I have learnt that when we pass the array name to sizeof, the name of the array does not decay to the pointer to base address. The code below verifies this fact by giving answer 10. #include <stdio.h> int main(){ int arr[10]; printf("Size of array is %d" , sizeof(arr)/sizeof(int)); return 0; } However when I run the code below, the answer comes 1. Irrespective of whether a dimension is written in prototype or not , the answer is 1. Why is it so ? #include <stdio.h> void dimension(int arr[]){ printf("Sizof array is %d" , sizeof(arr)/sizeof(int)); } int main(){ int arr[10]; dimension(arr);

c语言中，定义数组后可以用sizeof命令获取数组的长度（可容纳元素个数）； 如： { int data[5]; int length; length=sizeof(data)/sizeof(data[0]);//数组占内存总空间，除以单个元素占内存空间大小 printf("length of data[5]=%d",length);输出length of data[5]=5 } 但是通过传递数组名参数到子函数中，以获取数组长度是不可行的，如 int getLength(int[] a){ int length; length=sizeof(a)/sizeof(a[0]);//这样得到的结果永远是1 return length; } 因为，a是函数参数，到了本函数中，a是一个指针（地址，系统在本函数运行时，是不知道a锁表示的地址有多大的数据存储空间，这里只是告诉函数：一个数据存储空间首地址），所以sizeof(a)的结果是指针变量a占内存的大小，一般在32位机上是4个字节，a[0]是int类型，sizeof(a[0])也是4个字节，所以结果永远是1。 因此，获得数组长度，只能在数组定义所在的代码区中，采用以上方法，才可以达到效果 来源： https://www.cnblogs.com/dream-to-pku/p/11381223.html

Why can't I use sizeof() on simple structs? eg: private struct FloatShortPair { public float myFloat; public short myShort; }; int size = sizeof(FloatShortPair); //CS0233 error CS0233: 'FloatShortPair' does not have a predefined size, therefore sizeof can only be used in an unsafe context (consider using System.Runtime.InteropServices.Marshal.SizeOf) MSDN states: The sizeof operator can only be used for types that are compile-time constants. If you are getting this error, make sure that the size of the identifier can be determined at compile time. If it cannot, then use SizeOf instead of

一、文件头： sizeof(Elf32_Ehdr)=52 e_type： 类型---可重定位文件.o、可执行文件、共享目标文件.so e_ehsize： 文件头大小---52 e_machine： CPU平台属性，如Intel_x86； e_entry： 入口虚拟地址，但是可重定位文件没有； e_shoff： 段表偏移量， 即在ELF文件的位置；下面两个11*40=440表达段表长度 e_shnum： 段表描述符数量，即段表中段的个数：11 e_shentsize： 段表描述符大小，sizeof(Elf32_Shdr)=40 e_shstrndx： 段表字符串表所在的段在段表中的下标； 二、段表： sizeof(Elf32_Shdr)=40 sh_name： 段名，如.test/.data/.bss/.rodata/.comment, .rel.text/.rel.data, .symtab, .strtab/.shstrtab, sh_type： 类型---程序段、重定位表、符号表、字符串表 sh_flags： 标志位---可写、可执行、分配空间 sh_addr： 加载后在地址空间中的虚拟地址 sh_offset： 段偏移量，即该段该文件中的地址！！！！！！！！！！！！！！！！！ sh_size： 段大小 sh_link： 符号表的下标 sh_info： 作用于哪个段？ 三

I have a vector<set<char> > data structure (transactions database) and I want to know the size of it. When I use sizeof() with each set<char> the size is 24 in spite of the set contains 3, 4 or 5 chars. Later, when I use sizeof() with the vector<set<char> > the size is 12... I suppose this is not the way to know the size of a data structure. Any help? Thanks. You want vector::size() and set::size() . Assuming v is your vector, do this: size_t size = 0; for (vector<set<char> >::const_iterator cit = v.begin(); cit != v.end(); ++cit) { size += cit->size(); } sizeof() is giving you the in-memory

An argument in the comments section of this answer prompted me to ask this question. In the following code, bar points to a variable length array, so the sizeof is determined at runtime instead of compile time. int foo = 100; double (*bar)[foo]; The argument was about whether or not using sizeof evaluates its operand when the operand is a variable length array, making sizeof(*bar) undefined behavior when bar is not initialized. Is it undefined behavior to use sizeof(*bar) because I'm dereferencing an uninitialized pointer? Is the operand of sizeof actually evaluated when the type is a variable