WebMar 26, 2013 · This is fixed. You can't change the size of this array. The size N has to be a compile-time constant. T* arr = new T[N]; This declaration defines a T* with automatic storage duration. However, it also creates an array of size N with dynamic storage duration. Here, the size N doesn't need to be a compile-time constant. However, this doesn't help ... WebIf you mean a C-style array, then you can do something like: int a[7]; std::cout << "Length of array = " << (sizeof(a)/sizeof(*a)) << std::endl; This doesn't work on pointers (i.e. it won't work for either of the following):
[c++] How do I find the length of an array? - SyntaxFix
WebOct 23, 2024 · Instead of using "String" (Whatever that is in your case.), use a fixed array of char for each string. Like so: struct person { char name [32], /* string name has a length of 31 + 1 for NULL-termination */ char city [32], /* string city has a length of 31 + 1 for NULL-termination */ int age } Share Improve this answer Follow WebSep 23, 2014 · In standard C you can end a struct with an array of size 0 and then over allocate it to add a variable length dimension to the array: struct var { int a; int b []; } struct var * x=malloc (sizeof (var+27*sizeof (int))); How can you do that in C++ in a standard (portable) way? brushed stainless steel switch
sizeof single struct member in C - Stack Overflow
WebNov 5, 2010 · If you mean a C-style array, then you can do something like: int a [7]; std::cout << "Length of array = " << (sizeof (a)/sizeof (*a)) << std::endl; This doesn't work on pointers (i.e. it won't work for either of the following ): int *p = new int [7]; std::cout << "Length of array = " << (sizeof (p)/sizeof (*p)) << std::endl; or: WebIn C struct array elements must have a fixed size, so the char *theNames [] is not valid. Also you can not initialize a struct that way. In C arrays are static, i.e. one cannot change their size dynamically. A correct declaration of the struct would look like the following struct potNumber { int array [20]; char theName [10] [20]; }; Webcalculate it when it's still an array, and pass that size to any functions. same as above but using funky macro magic, something like: #define arrSz(a) (sizeof(a)/sizeof(*a)). create your own abstract data type which maintains the length as an item in a structure, so that you have a way of getting your Array.length(). brushed stainless steel sugar bowl