[英]c optarg atoi with no args
Consider the following code: 考虑以下代码:
int number;
while((w = getopt(argc, argv, "n:s:")) != -1) {
switch (w){
case 'n': {
opfile->filename = optarg;
}break;
case 's': {
number = atoi(optarg);
}break;
}
}
Now, when I leave both options or the option s
blank, for example I start my program with no command line args, then the number
variable still gets a random value. 现在,当我将两个选项或选项
s
都保留为空白时,例如,我在没有命令行参数的情况下启动程序,那么number
变量仍将获得随机值。
What am I missing here? 我在这里想念什么? Some if-statement in the case of
s
? 关于
s
一些if陈述? Specifically, I want to cover the case where the user doesn't assign a specific value/option to s
in the command line arguments. 具体来说,我想介绍用户未在命令行参数中为分配特定值/选项
s
情况。
When there is no 's' option passed to the program, the case 's'
branch is not executed at all, and nothing else sets number
to a value, which means that subsequent reads trigger undefined behavior. 当没有“ s”选项传递给程序时,
case 's'
分支根本不会执行,其他都不会将number
设置为值,这意味着后续的读取会触发未定义的行为。 (This is potentially much worse than just giving you a random value when you read from it later. It's a must-fix bug.) (这可能比以后读取时给您一个随机值要糟糕得多。这是一个必须修复的错误。)
But because nothing else touches number
, it will be enough to change 但是,因为没有别的东西触及
number
,所以改变就足够了
int number;
to 至
int number = 0;
or whatever else you want your default to be. 或其他任何您想要的默认值。
(By the way, you should really be using strtol
instead of atoi
, because atoi
ignores syntax errors.) (顺便说一句,您确实应该使用
strtol
而不是atoi
,因为atoi
忽略语法错误。)
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