我非常关心C的扼杀。 我是否需要确定最后的char,还是由它自己决定? 如果我不以人工的方式行事,那么当我试图推翻法典时,当我看看着1[257]号指示时,这并非无效。 我在解开一系列的雕像时有问题,因此我认为这是一个原因。
char string1[257], string2[257];
scanf("%s", &string2);
string1[257] = ;
strncpy(string1, string2, 257);
string1[257] = ; /* do I need to do that? */
是否绝对必要? <>No,因为在您打电话scanf时,strcpy(strncpy除外。 如果你在体力上填满了零件,就把零件复印给你。 这样做是否好? <>Nostrong>No really, 它确实无助于缓冲外流问题,因为这些功能将超越缓冲地带的规模。 那么,什么是最佳方法? 使用<代码>的++:载。
顺便提一句,如果您能够查阅<>条码>,则该编码将不受约束,因为您可重新查阅/撰写第258条<>第号<>在一系列大小257中的内容。 (零基指数)
封面字像>“Hello World >/code> 无效,但char 阵列并不自动终止。
始终采取的一般原则一是谨慎行事,并将<条码>>>>转让至扼杀结束,除非造成履约问题。 在这些情况下,我对图书馆的职能是我使用的额外谨慎。
仔细分配足够的记忆,比较以下编码线的影响:
char s1[3] = "abc";
char s2[4] = "abc";
char s3[] = "abc";
所有这三条法律系:,但第一个情况下,第四种无效性质有足够的记忆。 第1条不会像正常的扼杀一样,而是第2和第3条。 汇编者自动计算3分,你获得4个分配记忆。 如果你试图写字
s1[3] = ;
这种做法不明确,你重新撰写,以缅怀那些不属于1岁的人,并且会产生令人厌恶的影响,甚至可能破坏小鼠背后的信息,使其难以自由记忆。
A literal string like "foo
bar" is always translated to a const char literal[] with an additional zero byte at the end. (So the constant would have 8 bytes, the first being f and the last being zero).
但是,在<代码>strncpy后,你有权明确将最后的 by升至0。
正如Aurelio De Rosa所说,最后一个正确的指数是256个阵列<代码>[257]。
是的,你需要这样做。 并非所有职能都给你留下了独一无二的char子,而且正如我可以在一页读的那样,<条码>斯特凡·要求rc首的独一无二。
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