Answer from the C FAQ: The Prime 50 series used segment 07777, offset 0 for the null pointer, at least for PL/I. Later models used segment 0, offset 0 for null pointers in C, necessitating new instructions such as TCNP (Test C Null Pointer), evidently as a sop to all the extant poorly-written C code … Read more
It is always one in C99, section 6.5.3.4: When applied to an operand that has type char, unsigned char, or signed char, (or a qualified version thereof) the result is 1. Edit: not part of your question, but for interest from Harbison and Steele’s. C: A Reference Manual, Third Edition, Prentice Hall, 1991 (pre c99) … Read more
Difference between &str and str, when str is declared as char str[10]? Read sizeof Operator: 6.5.3.4 The sizeof operator, 1125: When you apply the sizeof operator to an array type, the result is the total number of bytes in the array. So, according to your declaration, sizeof(str2) gives the complete array size that is 10 … Read more
discussion on same subject “More specifically the integral promotions. In K&R C it was virtually (?) impossible to use a character value without it being promoted to int first, so making character constant int in the first place eliminated that step. There were and still are multi character constants such as ‘abcd’ or however many … Read more
There is no magic solution. C is not a reflective language. Objects don’t automatically know what they are. But you have many choices: Obviously, add a parameter Wrap the call in a macro and automatically add a parameter Use a more complex object. Define a structure which contains the dynamic array and also the size … Read more
Just use the sys.getsizeof function defined in the sys module. sys.getsizeof(object[, default]): Return the size of an object in bytes. The object can be any type of object. All built-in objects will return correct results, but this does not have to hold true for third-party extensions as it is implementation specific. Only the memory consumption … Read more
Yes it’s inherited from C. The function: void foo ( char a[100] ); Will have the parameter adjusted to be a pointer, and so becomes: void foo ( char * a ); If you want that the array type is preserved, you should pass in a reference to the array: void foo ( char (&a)[100] … Read more
An array-type is implicitly converted into pointer type when you pass it in to a function. So, void PrintSize(int p_someArray[10]) { printf(“%zu\n”, sizeof(p_someArray)); } and void PrintSize(int *p_someArray) { printf(“%zu\n”, sizeof(p_someArray)); } are equivalent. So what you get is the value of sizeof(int*)
No, you can’t. The compiler doesn’t know what the pointer is pointing to. There are tricks, like ending the array with a known out-of-band value and then counting the size up until that value, but that’s not using sizeof(). Another trick is the one mentioned by Zan, which is to stash the size somewhere. For … Read more
This is because of padding added to satisfy alignment constraints. Data structure alignment impacts both performance and correctness of programs: Mis-aligned access might be a hard error (often SIGBUS). Mis-aligned access might be a soft error. Either corrected in hardware, for a modest performance-degradation. Or corrected by emulation in software, for a severe performance-degradation. In … Read more