[英]Is this undefined behavior in C? (c=x) + (c==y)
I have a bit of sample code that is throwing this warning: 我有一些示例代码抛出此警告:
main.c: In function ‘getline_’:
main.c:30:32: warning: operation on ‘c’ may be undefined [-Wsequence-point]
In this particular exercise I was to avoid using the ||
在这个特定的练习中,我要避免使用||
and &&
operator, but this doesn't seem like it should produce undefined behavior. 和&&
运算符,但这似乎不应该产生未定义的行为。 The compiler message is just a warning, but I wanted to know for knowings sake. 编译器的消息只是一个警告,但我想知道knowings缘故。 Is this code actually going to produce undefined behavior? 这段代码实际上会产生未定义的行为吗?
24 int getline_( char s[], int limit)
25 {
26 int i, c;
27 i=0;
28 for( i=0; (i<limit-1) + ((c=getchar())!='\n') + (c!=EOF) == 3; i++){
29 s[i]=c;
30 }
31 if( c == '\n' ){
32 s[i]=c;
33 i++;
34 }
35 s[i]='\0';
36 return i;
37 }
It seems to work ok in my basic tests. 它似乎在我的基本测试中正常工作。
Edit: Updated title as per comment, thanks pst. 编辑:根据评论更新标题,谢谢pst。
This is unspecified behavior: 这是未指定的行为:
(i<limit-1) + ((c=getchar())!='\\n') + (c!=EOF) == 3
the order of evaluation of expressions between sequence points is unspecified in C. It is unspecified if the assignment to c
occurs before the equality check with EOF
. 在C中未指定序列点之间的表达式的评估顺序。如果在使用EOF
进行相等性检查之前发生对c
的赋值,则未指定。
In addition to the unspecified behavior, it is also undefined behavior because it violates the sequence points rules and particularly this one: 除了未指定的行为之外,它还是未定义的行为,因为它违反了序列点规则,特别是这个:
(C99, 6.5p2) "Furthermore, the prior value shall be read only to determine the value to be stored." (C99,6.5p2)“此外,先前的值应该是只读的,以确定要存储的值。”
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