The ISO C standard that came into being in 1999.
An array whose length is not known in compilation time. int testfun (int n) { int varr [n]; ... } it is ok in C99, invalid in earlier versions.
The latest ANSI standard is C99. See the attached link.
For the C99 standard, __STDC_VERSION__ is a mandatory macro.This macro was not specified in ISO/IEC9899:1990 and was specified as 199409L inISO/IEC 9899/AMD1:1995 and as 199901L in ISO/IEC9899:1999 for C99. The intention was that this would remain an integer constant of type long int that is increased with each revision of the International Standard.According to the most current International Standard, ISO/IEC 9899:201x , document N1539, it shall be defined by the implementation to be the integer constant 201ymmL
Standard C (C89 and C99 are the official standards) does not allow to define functions inside functions (known as nestedfunctions). However, some compilers, such as the free GCC allow nested functions as a language extension.
For C++ you can do this:#define member_offset(Struct, Field) (&(reinterpret_cast(0)->member))because the byte offset is usually the structure address + offset, so by setting the address to 0, you get 0 + offset == offset.AnswerYou should the "offsetof" macro. This is a standard macro that has been a part of the C standard for 20 years now (it's part of C89 and C99), so all C and C++ compilers should support it.#include size_t offsetof(type, member);For example:struct s { int x, y; };offsetof(struct s, y);
The c99 command is a wrapper program that actually calls 'cc'. This is the standard c compiler for Linux. Since other Unix based systems use a c99 command to call the compiler with the 1999 standards there is a similar command to do the same thing under Linux.
There is no "latest version of C". However, you might be referring to the latest version of the C Standard, which is officially C99. Mind you that some compilers do not implement all features of C99.
An array whose length is not known in compilation time. int testfun (int n) { int varr [n]; ... } it is ok in C99, invalid in earlier versions.
Using this you specify that two pointers can't point on the same address
The language doesn't have versions, but there are standards like C99 or C1X (pending).
The latest ANSI standard is C99. See the attached link.
32 keywords are present in C language. There are 44 keywords in C99.
C for Dummies, 2nd Edition covers ANSI C99, which has beensupersededby C11, which was ratified in 2011; the 11 in C11 is short for 2011, while 99 in C99 is short for 1999. However, even though there is a new version of C that is gaining wide support in popular compilers such as GCC and MinGW, the principles outlined in this book are still largely useful and relevant in the new language specification, which primarily focused on adding features and removing only features that were deemed excessively risky to allow to remain the language. Almost any program written in C99 will compile and run in C11. One can learn the techniques and features in C99 using this book, then migrate to C11. There are no known books that teach C11 directly, since it is simply a revision of C99, and is largely compatible with it.
C language doesn't have versions; the latest commonly used standard is ANSI C99.
For the C99 standard, __STDC_VERSION__ is a mandatory macro.This macro was not specified in ISO/IEC9899:1990 and was specified as 199409L inISO/IEC 9899/AMD1:1995 and as 199901L in ISO/IEC9899:1999 for C99. The intention was that this would remain an integer constant of type long int that is increased with each revision of the International Standard.According to the most current International Standard, ISO/IEC 9899:201x , document N1539, it shall be defined by the implementation to be the integer constant 201ymmL
According to the current C99 standard, "an expression is a sequence of operators and operands that specifies computation of a value, or that designates an object or a function, or that generates side effects, or that performs a combination thereof." (pg. 67)
Yes, see the C99 ISO standard §6.5.2.1/2. Assuming a pointer "a" into an allocated array, a[-1] is equivalent to *(a - 1). Naturally, if (a - 1) points to an unallocated memory location, you get undefined behavior.