blocks|key|2038053|text|他们是不对的。它们似乎是工作的，因为您要求scanf存储一个整数，存储它的地址是指针+p的地址。基本上，您将指针本身的存储视为整数的存储。同样，对于printf，您传递指针的地址(其中包含整数)，并要求printf将其读取为.一个整数。您甚至可以将第一行更改为|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|BOLD|entityRanges|data|2038054|float*+p;|code-block|syntax|javascript|2038055|而且它似乎仍然有效。最后，这是一个很好的例子，说明了为什么您应该避免类型不安全的C风格接口。|2038056|entityMap^0|L|5|17|1|22|6|2T|6|19|2D|0|0|0|1A|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]|$9|Y|A|Z|B|D]]|E|@]|F|$]]|$1|G|3|H|5|I|7|10|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|11|8|@$9|12|A|13|B|D]]|E|@]|F|$]]|$1|N|3|-4|5|6|7|14|8|@]|E|@]|F|$]]]|O|$]]

They are not right. They just seem to work, because you ask <code>scanf</code> to store an integer and the address where to store it is the address of the pointer <code>p</code>. You are basically treating the storage of the pointer itself as the storage of an integer. Likewise for <code>printf</code>, you pass the address of the pointer (which contains the integer) and ask <code>printf</code> to read it from there as... an integer. You could even change the first line to

<pre><code>float* p;
</code></pre>

and it would still seem to work. In the end, this is a good example of why you should avoid C-style interfaces which are not type-safe.

blocks|key|803444|text|这只有在指针与整数大小相同的情况下才能工作，因为基本上是将指针视为整数。也就是说，如果int是32位整数，指针void*是32位地址.|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|803445|它的写法是：|803446|int+p;+//+not+the+lack+of+the+*
scanf("%25d",&p);+//+this+gives+scanf+the+address+of+p
printf("%25d\n",p);+//+this+uses+p's+value|code-block|syntax|javascript|803447|它将实际使用p作为整数，而不是将其声明为指针并将其视为整数。|803448|entityMap^0|17|3|1J|5|0|0|0|6|1|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@$9|Y|A|Z|B|C]]|D|@]|E|$]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

This will only work as long as a pointer is the same size as an integer because you are basically treating the pointer as an integer. That is, if <code>int</code> is a 32-bit integer, and a pointer <code>void*</code> is a 32-bit address. 

The way it should be written:

<pre><code>int p; // not the lack of the *
scanf("%d",&amp;p); // this gives scanf the address of p
printf("%d\n",p); // this uses p's value
</code></pre>

Which will actually use <code>p</code> as an integer instead of declaring it as a pointer and treating it like an integer.

blocks|key|3179954|text|如果我能通过你的陈述来解释，那就是|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3179955|int+*p;+//Declaration+of+pointer+variable+p,+which+can+hold+the+address+of+integer+variable
scanf("%25d",&p);+//Getting+input,+will+be+stored+at+address+of+pointer+variable+p
printf("%25d\n",p);+//It+will+display+the+value+stored+at+&p|code-block|syntax|javascript|3179956|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If I will explain through your statements, then it will be

<pre><code>int *p; //Declaration of pointer variable p, which can hold the address of integer variable
scanf("%d",&amp;p); //Getting input, will be stored at address of pointer variable p
printf("%d\n",p); //It will display the value stored at &amp;p
</code></pre>

blocks|key|803454|text|上面的代码并不是真正正确的，它建立在相当多的假设和C结构之上。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|803455|第一个假设是指针和int具有相同的大小，这将破坏大多数(我所知道的)64位平台。然后，它使用类型不安全的接口(变量函数参数)传递int*的地址，就好像它是int的地址一样。scanf内部的代码将假设它正在写入一个int并对这些位.但是这种类型不是int*，即使在sizeof(int)+=+sizeof(int*)平台上，这种位模式也可能有意义，也可能没有意义。|offset|length|style|CODE|803456|entityMap^0|0|9|3|1S|4|25|3|2E|5|2Y|3|3F|4|3N|Q|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]|$D|T|E|U|F|G]|$D|V|E|W|F|G]|$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|I|$]]

The code above is not really correct, and it is built on top of a fair amount of assumptions and C constructs.

The first assumption is that a pointer and an <code>int</code> have the same size, which will break in most (all I know) 64bit platforms. It is then using a type unsafe interface (variadic function arguments) to pass the address of an <code>int*</code> as if it was the address of an <code>int</code>. The code inside the <code>scanf</code> will assume that it is writting to an <code>int</code> and write over the bits... but the type is not an <code>int*</code> and that bit pattern might or not make sense even in the platforms where the <code>sizeof(int) = sizeof(int*)</code>

blocks|key|3180031|text|那代码一点也不对。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3180032|一些(坏的)编译器可能无法检测到问题，因为printf和scanf都是各种函数。但是，例如，g%2B%2B会警告您，传递的类型与格式字符串中指定的类型不匹配。|offset|length|style|CODE|3180033|entityMap^0|0|L|6|S|5|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

That code is not right at all.

Some (bad) compilers may be unable to detect the problems because both <code>printf</code> and <code>scanf</code> are variadic functions... but for example g++ would warn you that the types passed don't match with the ones specified in the format strings.

<pre><code>int *p;
scanf("%d",&amp;p);
printf("%d\n",p);
</code></pre>

In my past understandings,the "p" is a address,but now it seems that the p is a simple variable.

I cant understand why these 3 lines code are right!!!

can you help me?

why these 3 lines code are right?About *p

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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int *p;scanf("%d",&p);printf("%d\n",p);在我过去的理解中，"p“是一个地址，但现在看来p是一个简单的变量。我不明白为什么这3行代码是正确的！你能帮帮我吗?

问为什么这3行代码是正确的?大约*p
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问为什么这3行代码是正确的?大约*pEN

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问为什么这3行代码是正确的?大约*p
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