blocks|key|47806|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|47807|const++char数组不必位于只读内存中.试图修改它是一种未定义的行为。任何事情都可能发生分段故障、修改、病毒激活、银行账户转移、磁盘擦除等。|ordered-list-item|offset|length|style|CODE|47808|47809|warnings+|47810|读取编译器|BOLD|47811|47812|<source>:9:20:+warning:+passing+'const+char+[99]'+to+parameter+of+type+'char+*'+discards+qualifiers+[-Wincompatible-pointer-types-discards-qualifiers]
++++char+*p=strtok(a,"-");|code-block|syntax|javascript|47813|47814|int+main()应该是int+main(void)|47815|47816|entityMap^0|0|0|B|0|0|0|2|3|0|0|0|0|0|R|0|A|D|E|0|0^^$0|@$1|2|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|10|7|@$D|11|E|12|F|G]]|8|@]|9|$]]|$1|H|3|-4|4|5|6|13|7|@]|8|@]|9|$]]|$1|I|3|J|4|5|6|14|7|@]|8|@]|9|$]]|$1|K|3|L|4|C|6|15|7|@$D|16|E|17|F|M]]|8|@]|9|$]]|$1|N|3|-4|4|5|6|18|7|@]|8|@]|9|$]]|$1|O|3|P|4|Q|6|19|7|@]|8|@]|9|$R|S]]|$1|T|3|-4|4|5|6|1A|7|@]|8|@]|9|$]]|$1|U|3|V|4|C|6|1B|7|@$D|1C|E|1D|F|M]|$D|1E|E|1F|F|G]|$D|1G|E|1H|F|G]]|8|@]|9|$]]|$1|W|3|-4|4|5|6|1I|7|@]|8|@]|9|$]]|$1|X|3|-4|4|5|6|1J|7|@]|8|@]|9|$]]]|Y|$]]

1.<code> const char</code> array does not have to be in read-only memory. Attempt to modify it is an Undefined Behaviour. Anything may happen segfault, modification, virus activation, bank account transfer, disk erase etc etc.
<ol start="2">
<li>Read the compiler warnings</li>
</ol>
<pre><code>&lt;source&gt;:9:20: warning: passing 'const char [99]' to parameter of type 'char *' discards qualifiers [-Wincompatible-pointer-types-discards-qualifiers]
 char *p=strtok(a,&quot;-&quot;);
</code></pre>
<ol start="3">
<li><code>int main()</code> should be <code>int main(void)</code></li>
</ol>

blocks|key|48436|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|48437|，为什么在这里修改a？？|blockquote|48438|48439|48440|strtok修改字符串。修改const对象会导致未定义的行为。示例程序的行为未定义。|offset|length|style|CODE|48441|实际上，const数组不会转换为指向非const的指针(没有强制转换)，因此程序的格式不正确。|48442|48443|其他问题：|48444|不包括声明strtok.|48445|int+main()在C.|unordered-list-item|48446|
中不符合标准的标题。|48447|entityMap^0|0|0|0|0|0|6|0|0|0|0|5|6|0|0|A|0|0^^$0|@$1|2|3|-4|4|5|6|Z|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|10|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|11|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|12|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|13|7|@$H|14|I|15|J|K]]|8|@]|9|$]]|$1|L|3|M|4|5|6|16|7|@]|8|@]|9|$]]|$1|N|3|-4|4|5|6|17|7|@]|8|@]|9|$]]|$1|O|3|P|4|5|6|18|7|@]|8|@]|9|$]]|$1|Q|3|R|4|5|6|19|7|@$H|1A|I|1B|J|K]]|8|@]|9|$]]|$1|S|3|T|4|U|6|1C|7|@$H|1D|I|1E|J|K]]|8|@]|9|$]]|$1|V|3|W|4|5|6|1F|7|@]|8|@]|9|$]]|$1|X|3|-4|4|5|6|1G|7|@]|8|@]|9|$]]]|Y|$]]

<blockquote>
why a is modified here??
</blockquote>
<code>strtok</code> modifies the string. Modifying const object results in undefined behaviour. The behaviour of the example program is undefined.
In fact, the const array doesn't convert to pointer to non-const (without cast), so the program is ill-formed.
<hr />
Other problems:
<ul>
<li>You don't include the header that declares <code>strtok</code>.</li>
<li><code>int main()</code> is non-standard in C.</li>
</ul>

blocks|key|48449|text|来自C标准(6.7.3类型限定符)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|48450|如果试图通过使用具有非const限定类型的lvalue来修改使用const限定类型定义的对象，则该行为是未定义的。|48451|48452|编译器至少应该发出一个警告，即strtok的调用从传递的参数表达式中丢弃限定符const。|offset|length|style|CODE|48453|在strtok的召唤中|48454|char+*p=strtok(a,"-");|code-block|syntax|javascript|48455|数组a被隐式转换为const+char+*类型的指针到其第一个元素，而函数的相应参数没有限定符const。|48456|entityMap^0|0|0|0|F|6|13|5|0|1|6|0|0|9|C|1B|5|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|W|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|X|8|@]|9|@]|A|$]]|$1|E|3|F|5|6|7|Y|8|@$G|Z|H|10|I|J]|$G|11|H|12|I|J]]|9|@]|A|$]]|$1|K|3|L|5|6|7|13|8|@$G|14|H|15|I|J]]|9|@]|A|$]]|$1|M|3|N|5|O|7|16|8|@]|9|@]|A|$P|Q]]|$1|R|3|S|5|6|7|17|8|@$G|18|H|19|I|J]|$G|1A|H|1B|I|J]]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]]|U|$]]

From the C Standard (6.7.3 Type qualifiers)
<blockquote>
6 If an attempt is made to modify an object defined with a
const-qualified type through use of an lvalue with non-const-qualified
type, the behavior is undefined.
</blockquote>
The compiler should issue at least a warning that the call of <code>strtok</code> discards the qualifier <code>const</code> from the passed argument expression.
In this call of <code>strtok</code>
<pre><code>char *p=strtok(a,&quot;-&quot;);
</code></pre>
the array a is implicitly converted to pointer of the type <code>const char *</code> to its first element while the corresponding parameter of the function does not have the qualifier <code>const</code>.

blocks|key|2428075|text|const的定义不是“计算机将阻止您修改对象”。C+2018+6.7.3+7中const的定义是：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2428076|2428077|如果试图通过使用具有非const限定类型的lvalue来修改使用const限定类型定义的对象，则该行为是未定义的…。|blockquote|2428078|2428079|2428080|因此，使用const定义对象并不会从计算机中创建不修改对象的承诺。这实际上是另一个方向上的承诺:这是你对计算机的承诺，你不会试图修改对象。这使编译器可以将对象放入标记为只读的内存中.|2428081|如果您违背了承诺，则该行为不受C标准的定义。对象可能在只读内存中，试图修改它将导致陷阱，并提醒您注意程序中的错误。或者对象可能在可修改的内存中，试图修改它将修改它。或者，通过程序优化，可能会发生其他行为。|2428082|C标准确实在这方面给了你一些帮助。当您将一个const+char+*传递给strtok+(它需要一个char+*+)时，编译器需要发出一个诊断消息。注意编译器报告的警告和错误，并使用它们来修复程序。更可取的做法是，使用编译器开关将警告提升为错误。(-Werror与GCC或Clang，/WX与Microsoft+C%2B%2B.)|2428083|entityMap^0|0|5|13|5|0|0|0|0|0|5|5|0|0|M|C|11|6|1E|6|3G|7|3Y|3|0^^$0|@$1|2|3|4|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|Z|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|10|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|11|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|N|3|O|5|6|7|15|8|@]|D|@]|E|$]]|$1|P|3|Q|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]|$9|1F|A|1G|B|C]]|D|@]|E|$]]|$1|R|3|-4|5|6|7|1H|8|@]|D|@]|E|$]]]|S|$]]

The definition of <code>const</code> is not “the computer will prevent you from modifying the object”. The definition of <code>const</code> in C 2018 6.7.3 7 is:
<blockquote>
If an attempt is made to modify an object defined with a const-qualified type through use of an lvalue with non-const-qualified type, the behavior is undefined…
</blockquote>
So defining an object with <code>const</code> does not create a promise from the computer to you that the object will not be modified. It is actually a promise in the other direction: It is a promise from you to the computer that you will not attempt to modify the object. This gives the compiler permission to put the object in memory that is marked read-only.
If you violate the promise, the behavior is not defined by the C standard. The object might be in read-only memory, and attempting to modify it will cause a trap and alert you to a bug in your program. Or the object might be in modifiable memory, and attempting to modify it will modify it. Or, with program optimization, other behaviors may occur.
The C standard does give you some help with this. When you pass a <code>const char *</code> to <code>strtok</code>, which expects a <code>char *</code>, the compiler is required to issue a diagnostic message. Pay attention to the warnings and errors the compiler reports and use them to fix your program. Preferable, use a compiler switch to elevate warnings to errors. (<code>-Werror</code> with GCC or Clang, <code>/WX</code> with Microsoft Visual C++.)

<pre><code>#include &lt;stdio.h&gt;

int main() {
 const char a[99]=&quot;hello-hi&quot;;
 printf(&quot;%s\n&quot;,a);
 char *p=strtok(a,&quot;-&quot;);
 printf(&quot;%s&quot;,a);
 return 0;
}
</code></pre>
output:
<pre><code>hello-hi
hello
</code></pre>
why a is modified here?? I made it const but still why it is modified??

why constant char array is modified here when strtok() is used

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。

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#include <stdio.h>int main() { const char a[99]="hello-hi"; printf("%s\n",a); char *p=strtok(a,"-"); printf("%s",a); return 0;}产出：hello-hihello为什...

问当使用strtok()时，为什么在这里修改常量char数组
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问当使用strtok()时，为什么在这里修改常量char数组EN