c89

问题 The code I'm working on is supposed to be possible to build for both hosted and freestanding environments, providing private implementations for some stdlib functions for the latter case. Can I reliably test this with just GCC on a normal workstation/build server? Compile for freestanding environment with GCC The "-ffreestanding" option looked promising, but it seems that it "only" disables built-ins and sets the STDC_HOSTED macro properly, it still provides all system headers. The option "

问题 Is the difference of two non-void pointer variables defined (per C99 and/or C++98) if they are both NULL valued? For instance, say I have a buffer structure that looks like this: struct buf { char *buf; char *pwrite; char *pread; } ex; Say, ex.buf points to an array or some malloc'ed memory. If my code always ensures that pwrite and pread point within that array or one past it, then I am fairly confident that ex.pwrite - ex.pread will always be defined. However, what if pwrite and pread are

问题 While using gcc , the code: register a = 3; static b = 3; it is allowed while using the -std=c89 -pedantic-errors flags, although there is a warning. However it receive an error with the -std=c99 -pedantic-errors flags. I wonder which section of the C89 standards allows the "implicit int" rule? 回答1: The section that allowed the implicit int rule in C89 would be section 3.5.2 Type specifiers which says ( emphasis mine ): int , signed , signed int , or no type specifiers Keith Thompson in the

问题 I am trying to get Visual Studio to enforce the ANSI C standard when compiling a project, but I can't get it to work. Any tips? I have read all the tutorials, I enabled the /Za option, and named my file as .c (not .cpp). However, the following program still builds successfully: #include <stdio.h> void main(void) { for (int i = 0; i < 10; i++) { } int f = 0; } But it shouldn't. It would have to be like this to respect the ANSI C standard: #include <stdio.h> void main(void) { int i; int f = 0;

问题 I am confused as to what ANSI specification says about changing a variable declared const can be legally modified through its address. Unfortunately I do not have access to C90 specification but got conflicting pointers: The keyword const doesn't turn a variable into a constant! A symbol with the const qualifier merely means that the symbol cannot be used for assignment. This makes the value re ad -onl y through that symbol; it does not prevent the value from being modified through some other

问题 Zero-length arrays are allowed in GNU C. and can be initialized thus struct line { int length; char contents[0]; }; struct line *thisline = (struct line *) malloc (sizeof (struct line) + this_length); thisline->length = this_length; Note: I am referring to this page here: http://gcc.gnu.org/onlinedocs/gcc/Zero-Length.html (provides a basic intro to variable length structures in C) It goes on to say: "In ISO C90, you would have to give contents a length of 1, which means either you waste space

问题 For example, suppose you have these variables: int i = 9; int j = 7; Depending on the implementation, the value of, (-i)/j , could be either –1 or –2 . How is it possible to get these two different results? 回答1: Surprisingly the result is implementation defined in C89: ANSI draft § 3.3.5 When integers are divided and the division is inexact , if both operands are positive the result of the / operator is the largest integer less than the algebraic quotient and the result of the % operator is

问题 i'm trying to develop an application (written in ANSI C) for an OpenWrt router using libuci. I've read this useful post: How to find out if the eth0 mode is static or dhcp? and i've develop a piece of my application that is able to read network data (in this case i read if ppp is enabled) using uci library. char path[]="network.ppp.enabled"; struct uci_ptr ptr; struct uci_context *c = uci_alloc_context(); if(!c) return; if (strcmp(typeCmd, "GET") == 0){ if ((uci_lookup_ptr(c, &ptr, path, true

问题 I read with interest the post "How universally is C99 supported ?". One of the comments therein points that Microsoft doesn't support C99. But the comment symbol // works with VS 2008 and this symbol is in C99. I have two questions: To what extent VS 2008 support C99? Is it ok in the same code to mix C89 and C99 syntax together? So if I write my code in C89 and then place a comment //. This means that I have mixed-coding. So what does the compiler do in such a case? Check my code first with

问题 Original Question (please see "Edit: Updated scenario") This question might be a duplicate in one or another way to a huge collection of questions around undefined behavior of pointers to objects that gone out of scope etc. But all questions I found here to this topic are mostly specialized use cases. So I would like to spin the question upside down, not asking if something is forbidden, but WHAT exactly is allowed? To have a possible scenario: You have a function, that takes a pointer - you