blocks|key|4385289|text|阅读和获取与1,000,000,000表单的一些关联可能会更简单。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4385290|从技术角度看，我猜直接数和乘法没有区别。无论如何，编译器都会将其生成为常量十亿数字。|4385291|如果您谈到objective-c，那么1000%5E3将无法工作，因为没有这样pow语法(它是xor)。相反，可以使用pow()函数。但在这种情况下，它将不是最优的，它将是一个运行时函数调用，而不是编译器生成的常量。|4385292|entityMap^0|6|D|0|0|J|6|1L|5|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|P|8|@$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|K|$]]

Might be simpler to read and get some associations with the <code>1,000,000,000</code> form. 

From technical aspect I guess there is no difference between the direct number or multiplication. The compiler will generate it as constant billion number anyway.

If you speak about objective-c, then <code>1000^3</code> won't work because there is no such syntax for pow (it is xor). Instead, <code>pow()</code> function can be used. But in that case, it will not be optimal, it will be a runtime function call not a compiler generated constant.

blocks|key|4385319|text|以乘法方式声明常量的一个原因是提高可读性，同时不影响运行时性能。此外，还表明作者对数字的思考是乘法的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385320|请考虑以下内容：|4385321|double+memoryBytes+=+1024+*+1024+*+1024;|code-block|syntax|javascript|4385322|它显然比：|4385323|double+memoryBytes+=+1073741824;|4385324|因为后者乍一看并不是1024的第三次幂。|4385325|正如Amin+Negm-Awad提到的，%5E运算符是二元XOR。许多语言缺少内置的编译时求幂运算符，因此需要乘法运算。|offset|length|style|CODE|4385326|entityMap^0|0|0|0|0|0|0|K|1|R|3|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Y|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@$Q|13|R|14|S|T]|$Q|15|R|16|S|T]]|9|@]|A|$]]|$1|U|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|V|$]]

One reason to declare constants in a multiplicative way is to improve readability, while the run-time performance is not affected.
Also, to indicate that the writer was thinking in a multiplicative manner about the number.
Consider this:
<pre class="lang-c prettyprint-override"><code>double memoryBytes = 1024 * 1024 * 1024;
</code></pre>
It's clearly better than:
<pre class="lang-c prettyprint-override"><code>double memoryBytes = 1073741824;
</code></pre>
as the latter doesn't look, at first glance, the third power of 1024.
As Amin Negm-Awad mentioned, the <code>^</code> operator is the binary <code>XOR</code>. Many languages lack the built-in, compile-time exponentiation operator, hence the multiplication.

blocks|key|4385925|text|为了可读性。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385926|将逗号和空格放在零(1+000+000+000或1,000,000,000)之间会产生语法错误，并且在代码中使用1000000000会很难看到到底有多少个零。|offset|length|style|CODE|4385927|1000*1000*1000清楚地表明它是10%5E9，因为我们的眼睛可以更容易地处理块。而且，没有运行时开销，因为编译器会用常量1000000000替换它。|4385928|entityMap^0|0|A|D|O|D|1K|A|0|0|E|1R|A|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|M|8|@$D|N|E|O|F|G]|$D|P|E|Q|F|G]|$D|R|E|S|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|T|8|@$D|U|E|V|F|G]|$D|W|E|X|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|A|$]]]|K|$]]

For readability. 

Placing commas and spaces between the zeros (<code>1 000 000 000</code> or <code>1,000,000,000</code>) would produce a syntax error, and having <code>1000000000</code> in the code makes it hard to see exactly how many zeros are there. 

<code>1000*1000*1000</code> makes it apparent that it's 10^9, because our eyes can process the chunks more easily. Also, there's no runtime cost, because the compiler will replace it with the constant <code>1000000000</code>.

blocks|key|4385396|text|为了说明原因，请考虑以下测试程序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385397|$+cat+comma-expr.c+&&+gcc+-o+comma-expr+comma-expr.c+&&+./comma-expr
#include+<stdio.h>

#define+BILLION1+(1,000,000,000)
#define+BILLION2+(1000%5E3)

int+main()
{
++++++++printf("%25d,+%25d\n",+BILLION1,+BILLION2);
}
0,+1003
$|code-block|syntax|javascript|4385398|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|$]]

To illustrate the reasons consider the following test program:

<pre><code>$ cat comma-expr.c &amp;&amp; gcc -o comma-expr comma-expr.c &amp;&amp; ./comma-expr
#include &lt;stdio.h&gt;

#define BILLION1 (1,000,000,000)
#define BILLION2 (1000^3)

int main()
{
 printf("%d, %d\n", BILLION1, BILLION2);
}
0, 1003
$
</code></pre>

blocks|key|4385444|text|在C中实现十进制数类似效果的另一种方法是使用文字浮点表示法--只要双精度数可以表示您想要的数字而不损失任何精度。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4385445|IEEE754+64位double可以表示任何非负整数<=+2%5E53。通常，long+double+(80或128位)可以走得更远。转换将在编译时完成，因此没有运行时开销，如果存在意外的精度损失，并且您有一个好的编译器，您可能会收到警告。|4385446|long+lots_of_secs+=+1e9;|code-block|syntax|javascript|4385447|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Another way to achieve a similar effect in C for decimal numbers is to use literal floating point notation -- so long as a double can represent the number you want without any loss of precision.
IEEE 754 64-bit double can represent any non-negative integer &lt;= 2^53 without problem. Typically, long double (80 or 128 bits) can go even further than that. The conversions will be done at compile time, so there is no runtime overhead and you will likely get warnings if there is an unexpected loss of precision and you have a good compiler.
<pre class="lang-c prettyprint-override"><code>long lots_of_secs = 1e9;
</code></pre>

In code created by Apple, there is this line:
<pre class="lang-c prettyprint-override"><code>CMTimeMakeWithSeconds( newDurationSeconds, 1000*1000*1000 )
</code></pre>
Is there any reason to express <code>1,000,000,000</code> as <code>1000*1000*1000</code>?
Why not <code>1000^3</code> for that matter?

Why write 1,000,000,000 as 1000*1000*1000 in C?

编译器

在Apple创建的代码中，有这样一行：CMTimeMakeWithSeconds( newDurationSeconds, 1000*1000*1000 )有理由将1,000,000,000表达为1000*1000*1000吗为什么不使用1000^3呢？

问为什么用C写1,000,000,000为100010001000？
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