blocks|key|3586829|text|scanf返回成功转换的项目数.或者有错误的EOF。因此，用合理的方式对条件进行编码。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3586830|scanfresult+=+scanf(...);
while+(scanfresult+!=+EOF)+/*+while+scanf+didn't+error+*/
while+(scanfresult+==+1)+/*+while+scanf+performed+1+assignment+*/
while+(scanfresult+>+2)+/*+while+scanf+performed+3+or+more+assignments+*/|code-block|syntax|javascript|3586831|人为实例|3586832|scanfresult+=+scanf("%25d",+&a);
/*+type+"forty+two"+*/
if+(scanfresult+!=+EOF)+/*+not+scanf+error;+runs,+but+`a`+hasn't+been+assigned+*/;
if+(scanfresult+!=+1)+/*+`a`+hasn't+been+assigned+*/;|3586833|3586834|编辑:添加了另一个人为的示例|BOLD|3586835|int+a[5],+b[5];
printf("Enter+up+to+5+pairs+of+numbers\n");
scanfresult+=+scanf("%25d%25d%25d%25d%25d%25d%25d%25d%25d%25d",+a%2B0,b%2B0,a%2B1,b%2B1,a%2B2,b%2B2,a%2B3,b%2B3,a%2B4,b%2B4);
switch+(scanfresult)+{
case+EOF:+assert(0+&&+"this+didn't+happen");+break;
case+1:+case+3:+case+5:+case+7:+case+9:
++++printf("I+said+**pairs+of+numbers**\n");
++++break;
case+0:
++++printf("What+am+I+supposed+to+do+with+no+numbers?\n");
++++break;
default:
++++pairs+=+scanfresult+/+2;
++++dealwithpairs(a,+b,+pairs);
++++break;
}|3586836|entityMap^0|0|5|0|0|0|0|0|0|E|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|12|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|13|8|@$9|14|A|15|B|R]]|D|@]|E|$]]|$1|S|3|T|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|V|$]]

<code>scanf</code> returns the number of items succesfully converted ... or EOF on error. So code the condition the way it makes sense.

<pre><code>scanfresult = scanf(...);
while (scanfresult != EOF) /* while scanf didn't error */
while (scanfresult == 1) /* while scanf performed 1 assignment */
while (scanfresult &gt; 2) /* while scanf performed 3 or more assignments */
</code></pre>

Contrived example

<pre><code>scanfresult = scanf("%d", &amp;a);
/* type "forty two" */
if (scanfresult != EOF) /* not scanf error; runs, but `a` hasn't been assigned */;
if (scanfresult != 1) /* `a` hasn't been assigned */;
</code></pre>

<hr>

Edit: added another more contrived example

<pre><code>int a[5], b[5];
printf("Enter up to 5 pairs of numbers\n");
scanfresult = scanf("%d%d%d%d%d%d%d%d%d%d", a+0,b+0,a+1,b+1,a+2,b+2,a+3,b+3,a+4,b+4);
switch (scanfresult) {
case EOF: assert(0 &amp;&amp; "this didn't happen"); break;
case 1: case 3: case 5: case 7: case 9:
 printf("I said **pairs of numbers**\n");
 break;
case 0:
 printf("What am I supposed to do with no numbers?\n");
 break;
default:
 pairs = scanfresult / 2;
 dealwithpairs(a, b, pairs);
 break;
}
</code></pre>

blocks|key|3629087|text|取决于您想要对格式错误的输入做什么-如果您的扫描模式不匹配，您可以得到0返回。因此，如果您在循环之外处理这种情况(例如，如果您将其视为输入错误)，则与1进行比较(或者您的scanf调用中有多少项)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3629088|entityMap^0|Z|1|23|1|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Depends what you want to do with malformed input - if your scan pattern isn't matched, you can get <code>0</code> returned. So if you handle that case outside the loop (for example if you treat it the same as an input error), then compare with <code>1</code> (or however many items there are in your scanf call).

blocks|key|3629121|text|来自http://www.cplusplus.com/reference/clibrary/cstdio/scanf/|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|3629122|3629123|在成功的情况下，该函数将返回已连续读取的项目数。如果出现匹配失败，此计数可以匹配预期的读数数，甚至更少，甚至为零。如果在成功读取任何数据之前输入失败，则返回EOF。|blockquote|3629124|3629125|3629126|确保读取预期项数的唯一方法是将返回值与该数字进行比较。|3629127|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cplusplus.com/reference/clibrary/cstdio/scanf/^0|2|1L|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@$A|U|B|V|1|W]]|C|$]]|$1|D|3|-4|5|6|7|X|8|@]|9|@]|C|$]]|$1|E|3|F|5|G|7|Y|8|@]|9|@]|C|$]]|$1|H|3|-4|5|6|7|Z|8|@]|9|@]|C|$]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]|$1|J|3|K|5|6|7|11|8|@]|9|@]|C|$]]|$1|L|3|-4|5|6|7|12|8|@]|9|@]|C|$]]]|M|$N|$5|O|P|Q|C|$R|S]]]]

From <a href="http://www.cplusplus.com/reference/clibrary/cstdio/scanf/" rel="nofollow">http://www.cplusplus.com/reference/clibrary/cstdio/scanf/</a>

<blockquote>
 On success, the function returns the
 number of items succesfully read. This
 count can match the expected number of
 readings or fewer, even zero, if a
 matching failure happens. In the case
 of an input failure before any data
 could be successfully read, EOF is
 returned.
</blockquote>

The only way to be sure that you read the number of items intended is to compare the return value to that number.

Ceteris paribus (well formed data, good buffering practices and what not), is there a reason why I prefer to loop while the return of <code>scanf</code> is 1, rather than <code>!EOF</code>? I may have read this somewhere, or whatever, but I may have it wrong as well. What do other people think?

While scanf!=EOF or scanf==1?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

Ceteris (格式良好的数据、良好的缓冲实践等等)，为什么我喜欢在返回scanf时循环，而不是!EOF呢？我可能在某个地方读过这篇文章，或者别的什么，但我也可能看错了。其他人怎么想？

问而扫描！=EOF或scanf==1？
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问而扫描！=EOF或scanf==1？EN