map() 返回 LIST 时出现语法错误 [英] Syntax error when map() returns LIST

问题描述

这行得通，

print map { $_." x" => $_ } 1..5;
print map { ("$_ x" => $_) } 1..5;
print map { ("$_ x") => $_ } 1..5;

但这会引发语法错误，

print map { "$_ x" => $_ } 1..5;

这是记录的错误，未记录的错误，还是我不明白为什么不应该编译?

为什么 perl 认为这应该是 map EXPR, LIST 而不是 map BLOCK LIST

Why perl thinks this should be map EXPR, LIST instead of map BLOCK LIST

推荐答案

为什么 perl 认为这应该是 map EXPR, LIST 而不是 map BLOCK LIST?

相关部分代码在toke.c，Perl的词法分析器(以下来自Perl 5.22.0):

The relevant section of code is in toke.c, Perl's lexer (the below is from Perl 5.22.0):

/* This hack serves to disambiguate a pair of curlies
 * as being a block or an anon hash.  Normally, expectation
 * determines that, but in cases where we're not in a
 * position to expect anything in particular (like inside
 * eval"") we have to resolve the ambiguity.  This code
 * covers the case where the first term in the curlies is a
 * quoted string.  Most other cases need to be explicitly
 * disambiguated by prepending a "+" before the opening
 * curly in order to force resolution as an anon hash.
 *
 * XXX should probably propagate the outer expectation
 * into eval"" to rely less on this hack, but that could
 * potentially break current behavior of eval"".
 * GSAR 97-07-21
 */
t = s;
if (*s == '\'' || *s == '"' || *s == '`') {
    /* common case: get past first string, handling escapes */
    for (t++; t < PL_bufend && *t != *s;)
        if (*t++ == '\\')
            t++;
    t++;
}
else if (*s == 'q') {
    if (++t < PL_bufend
        && (!isWORDCHAR(*t)
            || ((*t == 'q' || *t == 'x') && ++t < PL_bufend
                && !isWORDCHAR(*t))))
    {   
        /* skip q//-like construct */
        const char *tmps;
        char open, close, term;
        I32 brackets = 1;

        while (t < PL_bufend && isSPACE(*t))
            t++;
        /* check for q => */
        if (t+1 < PL_bufend && t[0] == '=' && t[1] == '>') {
            OPERATOR(HASHBRACK);
        }
        term = *t;
        open = term;
        if (term && (tmps = strchr("([{< )]}> )]}>",term)))
            term = tmps[5];
        close = term;
        if (open == close)
            for (t++; t < PL_bufend; t++) {
                if (*t == '\\' && t+1 < PL_bufend && open != '\\')
                    t++;
                else if (*t == open)
                    break;
            }
        else {
            for (t++; t < PL_bufend; t++) {
                if (*t == '\\' && t+1 < PL_bufend)
                    t++;
                else if (*t == close && --brackets <= 0)
                    break;
                else if (*t == open)
                    brackets++;
            }
        }
        t++;
    }
    else
        /* skip plain q word */
        while (t < PL_bufend && isWORDCHAR_lazy_if(t,UTF))
             t += UTF8SKIP(t);
}
else if (isWORDCHAR_lazy_if(t,UTF)) {
    t += UTF8SKIP(t);
    while (t < PL_bufend && isWORDCHAR_lazy_if(t,UTF))
         t += UTF8SKIP(t);
}
while (t < PL_bufend && isSPACE(*t))
    t++;
/* if comma follows first term, call it an anon hash */
/* XXX it could be a comma expression with loop modifiers */
if (t < PL_bufend && ((*t == ',' && (*s == 'q' || !isLOWER(*s)))
                   || (*t == '=' && t[1] == '>')))
    OPERATOR(HASHBRACK);
if (PL_expect == XREF)
{
  block_expectation:
    /* If there is an opening brace or 'sub:', treat it
       as a term to make ${{...}}{k} and &{sub:attr...}
       dwim.  Otherwise, treat it as a statement, so
       map {no strict; ...} works.
     */
    s = skipspace(s);
    if (*s == '{') {
        PL_expect = XTERM;
        break;
    }
    if (strnEQ(s, "sub", 3)) {
        d = s + 3;
        d = skipspace(d);
        if (*d == ':') {
            PL_expect = XTERM;
            break;
        }
    }
    PL_expect = XSTATE;
}
else {
    PL_lex_brackstack[PL_lex_brackets-1] = XSTATE;
    PL_expect = XSTATE;
}

---

说明

如果开头卷曲后的第一个术语是字符串(由 '、" 或 ` 分隔)或裸字开头用大写字母，后面的术语是 , 或 =>，卷曲被视为匿名散列的开头(这就是 OPERATOR(HASHBRACK)); 表示).

---

Explanation

If the first term after the opening curly is a string (delimited by ', ", or `) or a bareword beginning with a capital letter, and the following term is , or =>, the curly is treated as the beginning of an anonymous hash (that's what OPERATOR(HASHBRACK); means).

其他情况对我来说有点难以理解.我通过 gdb 运行了以下程序:

The other cases are a little harder for me to understand. I ran the following program through gdb:

{ (x => 1) }

并在最后的 else 块中结束:

and ended up in the final else block:

else {
    PL_lex_brackstack[PL_lex_brackets-1] = XSTATE;
    PL_expect = XSTATE;
}

可以这么说，执行路径明显不同；它最终被解析为一个块.

Suffice it to say, the execution path is clearly different; it ends up being parsed as a block.

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