blocks|key|722424|text|决定您是否必须释放的不是变量的类型，而是它驻留在内存的哪一部分。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|722425|如果使用malloc或相关函数分配内存，则内存将位于堆上，因此必须对它们调用free。|offset|length|style|CODE|722426|在您的示例中，这些字符串是在静态存储器中创建的(不是按照其中一个注释中的建议从堆栈中创建的，否则在函数返回时不能使用它，因为它们本来会被覆盖)。|722427|这些字符串被编译成运行时二进制本身(您可以运行strings+<your+binary>并看到它们)。一旦程序被加载，内存中的一个特殊区域将被保留给这样的常量数据，因此您不需要释放它。|722428|另外，您必须从右到左读取C/C%2B%2B类型，所以类型const+char*意味着您有一个指向常量字符串的指针。这意味着您不能修改它所指向的字符串，但它没有说明如何分配这个指针后面的内存。它可能在堆上，在堆栈上，或者在静态内存上，只是类型不会告诉您。|722429|entityMap|0|LINK|mutability|MUTABLE|url|https://en.wikipedia.org/wiki/Memory_management#HEAP|1|https://en.wikipedia.org/wiki/Static_variable|2|https://en.wikipedia.org/wiki/Local_variable^0|0|4|6|12|4|Q|1|0|0|E|5|1|0|N|L|0|O|B|2R|2|2|0^^$0|@$1|2|3|4|5|6|7|Z|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|10|8|@$D|11|E|12|F|G]|$D|13|E|14|F|G]]|9|@$D|15|E|16|1|17]]|A|$]]|$1|H|3|I|5|6|7|18|8|@]|9|@$D|19|E|1A|1|1B]]|A|$]]|$1|J|3|K|5|6|7|1C|8|@$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|L|3|M|5|6|7|1F|8|@$D|1G|E|1H|F|G]]|9|@$D|1I|E|1J|1|1K]]|A|$]]|$1|N|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|O|$P|$5|Q|R|S|A|$T|U]]|V|$5|Q|R|S|A|$T|W]]|X|$5|Q|R|S|A|$T|Y]]]]

What will determine if you have to free or not is not the type of the variable, but on which part of memory it resides.
If you allocate memory using <code>malloc</code> or a related function, it will live on the <a href="https://en.wikipedia.org/wiki/Memory_management#HEAP" rel="nofollow noreferrer">heap</a>, thus you have to call <code>free</code> on them.
In your example, these strings are created in <a href="https://en.wikipedia.org/wiki/Static_variable" rel="nofollow noreferrer">static memory</a> (not from the stack as suggested in one of the comments, otherwise you could not use it once the function returns, as they would have been overwritten).
These strings are compiled into the run-time binary itself (you can run <code>strings &lt;your binary&gt;</code> and you will see them. Once the program is loaded, a special area of the memory is reserved for such constant data, thus you don't need to free it.
As a side note, you have to read the C/C++ types from right-to-left, so
type <code>const char*</code> means is that you have a pointer to a constant string. That means that you cannot modify the string pointed by it, but it does not say anything about how the memory behind this pointer was allocated. It could be on the heap, on the <a href="https://en.wikipedia.org/wiki/Local_variable" rel="nofollow noreferrer">stack</a>, or on static memory, just the type will not tell you.

blocks|key|123012|text|结论是否定的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|123013|原因是free()应该用作malloc()、calloc()或realloc()的合作伙伴。或者说，那些alloc函数只返回一个指针，那么它是否可以由free()回收。|123014|一个附加的原因是，您在代码中编写的所有字符串都分配在.rodata节中，这意味着这些字符串加载在内存空间中，您不能编辑而只能读取，所以即使存在释放它们的方法，您仍然不能使用内存空间。|123015|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|H|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|I|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|J|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|G|$]]

The conclusion is NO.
The reason is free() should be used as the partner of malloc(), calloc() or realloc(). Or to say only a pointer is returned by those alloc function, could it be recycled by free().
And an addtional reason is all those strings you write in your code is allocated in section .rodata, that means those strings are loaded in a memory space you cannot edit but only read, so even there is a method to free them, you still cannot util the memory space.

blocks|key|123061|text|if/else梯子过于复杂，容易出现“复制/粘贴”错误。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|123062|似乎您需要一个“查找表”来将一个工作日的名称转换为一个不同的字符串。|123063|const+char+*getOrder+(const+char+*day)+{
++++char+*LUT[][2]+=+{
++++{+"Monday",+"first",+},
++++{+"Tuesday",+"second",+},
++++{+"Wednesday",+"third",+},
++++{+"Thursday",+"forth",+},
++++{+"Friday",+"fifth",+},
++++{+"Saturday",+"sixth",+},
++++{+"Sunday",+"seventh",+},
++++};
++++
++++for(+i+=+0;+i+<+sizeof(LUT)/sizeof(LUT[0]);+i%2B%2B+)
++++++++if(+strncmp(day,+LUT[i][0],+strlen(+LUT[i][0]+)+)+==+0+)
++++++++++++return+LUT[+i+][+1+];

++++return+"NOT+A+DAY+NAME";
}|code-block|syntax|javascript|123064|它还在反复搜寻..。|123065|下面是一个很小的散列函数，用于快速将字符串日期名称转换为整数值(大小写不敏感)。玩玩它。也许你会试着把这个调整成合理的平日名称。|123066|//+Convert+a+day+name+(or+3+letter+abbrev.)+to+index+(1-7,+Sun=1).+Beware+false+positives!
static+int+wkdayOrd(+char+cp[]+)+{+return+"65013427"[*cp/2+%2B+~cp[1]+&+0x7]+&+0x7;+}|123067|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|F|7|V|8|@]|9|@]|A|$G|H]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

The if/else ladder is overly complicated and prone to 'copy/paste' errors.
It seems you want a &quot;look up table&quot; to translate a weekday name to a different string.
<pre><code>const char *getOrder (const char *day) {
 char *LUT[][2] = {
 { &quot;Monday&quot;, &quot;first&quot;, },
 { &quot;Tuesday&quot;, &quot;second&quot;, },
 { &quot;Wednesday&quot;, &quot;third&quot;, },
 { &quot;Thursday&quot;, &quot;forth&quot;, },
 { &quot;Friday&quot;, &quot;fifth&quot;, },
 { &quot;Saturday&quot;, &quot;sixth&quot;, },
 { &quot;Sunday&quot;, &quot;seventh&quot;, },
 };
 
 for( i = 0; i &lt; sizeof(LUT)/sizeof(LUT[0]); i++ )
 if( strncmp(day, LUT[i][0], strlen( LUT[i][0] ) ) == 0 )
 return LUT[ i ][ 1 ];

 return &quot;NOT A DAY NAME&quot;;
}
</code></pre>
It's still iteratively hunting...
Here is a tiny hash function to quickly translate string day names to an integer value (case insensitive). Play with it. Perhaps you will try to adapt this to go straight to the plausible weekday name.
<pre><code>// Convert a day name (or 3 letter abbrev.) to index (1-7, Sun=1). Beware false positives!
static int wkdayOrd( char cp[] ) { return &quot;65013427&quot;[*cp/2 + ~cp[1] &amp; 0x7] &amp; 0x7; }
</code></pre>

Please consider the following code. I am asking if this is correct, is there something missing. Is there any other way to make it.
<pre><code>#include &lt;stdio.h&gt;
#include &lt;string.h&gt;
const char *getOrder (const char *day)
{
 if (strncmp(day, &quot;Monday&quot;, 7) == 0) {
 return &quot;first&quot;;
 } else if (strncmp(day, &quot;Tuesday&quot;, 7) == 0) {
 return &quot;second&quot;;
 } else if (strncmp(day, &quot;Wednesday&quot;, 9) == 0) {
 return &quot;third&quot;;
 } else if (strncmp(day, &quot;Thursday&quot;, 8) == 0) {
 return &quot;forth&quot;;
 } else if (strncmp(day, &quot;Friday&quot;, 6) == 0) {
 return &quot;fifth&quot;;
 } else if (strncmp(day, &quot;Saturday&quot;, 8) == 0) {
 return &quot;sixth&quot;;
 } else if (strncmp(day, &quot;Sunday&quot;, 6) == 0) {
 return &quot;seventh&quot;;
 } else
 return NULL;
}
int main()
{
 const char* str = NULL;
 str = getOrder (&quot;Monday&quot;);
 printf(&quot;str : %s\n&quot;, str);
 return 0;
}
</code></pre>

The const char* should it be freed in C?

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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请考虑以下代码。我要问的是，如果这是正确的，是否有什么遗漏。还有别的办法可以做吗。#include <stdio.h>#include <string.h>const char *getOrder (const char *day){ if (strncmp(day, "Monday", 7) == 0) { ...

问const char*应该在C中释放吗？
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