blocks|key|1173226|text|#include+<climits>
#include+<iostream>
using+namespace+std;

int+main()+{
++cout+<<+INT_MAX+<<+endl;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1173227|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>#include &lt;climits&gt;
#include &lt;iostream&gt;
using namespace std;

int main() {
 cout &lt;&lt; INT_MAX &lt;&lt; endl;
}
</code></pre>

blocks|key|1388645|text|为什么不写一段代码，比如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1388646|int++max_neg+=+~(1+<<+31);
int++all_ones+=+-1;
int+max_pos+=+all_ones+&+max_neg;|code-block|syntax|javascript|1388647|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|$]]

Why not write a piece of code like:

<pre><code>int max_neg = ~(1 &lt;&lt; 31);
int all_ones = -1;
int max_pos = all_ones &amp; max_neg;
</code></pre>

blocks|key|1173317|text|下面是我用来获取带符号整数的最大值的宏，它与所使用的带符号整数类型的大小无关，对于这个宏，gcc+-Woverflow不会抱怨|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1173318|#define+SIGNED_MAX(x)+(~(-1+<<+(sizeof(x)+*+8+-+1)))

int+a+=+SIGNED_MAX(a);
long+b+=+SIGNED_MAX(b);
char+c+=+SIGNED_MAX(c);+/*+if+char+is+signed+for+this+target+*/
short+d+=+SIGNED_MAX(d);
long+long+e+=+SIGNED_MAX(e);|code-block|syntax|javascript|1173319|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|$]]

Here is a macro I use to get the maximum value for signed integers, which is independent of the size of the signed integer type used, and for which gcc -Woverflow won't complain

<pre><code>#define SIGNED_MAX(x) (~(-1 &lt;&lt; (sizeof(x) * 8 - 1)))

int a = SIGNED_MAX(a);
long b = SIGNED_MAX(b);
char c = SIGNED_MAX(c); /* if char is signed for this target */
short d = SIGNED_MAX(d);
long long e = SIGNED_MAX(e);
</code></pre>

blocks|key|1388787|text|那(1+<<+(8*sizeof(int)-2))+-+1+%2B+(1+<<+(8*sizeof(int)-2))呢？这与2%5E(8*sizeof(int)-2)+-+1+%2B+2%5E(8*sizeof(int)-2)相同。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1388788|如果是sizeof(int)+=+4++=>++2%5E(8*4-2)+-+1+%2B+2%5E(8*4-2)++=++2%5E30+-+1+%2B+20%5E30++=++(2%5E32)/2+-+1+[max+signed+int+of+4+bytes]。|1388789|您不能使用2*(1+<<+(8*sizeof(int)-2))+-+1，因为它会溢出，但(1+<<+(8*sizeof(int)-2))+-+1+%2B+(1+<<+(8*sizeof(int)-2))可以工作。|1388790|entityMap^0|1|1J|1O|19|0|3|34|0|5|U|18|1J|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|T|8|@$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|K|$]]

What about <code>(1 &lt;&lt; (8*sizeof(int)-2)) - 1 + (1 &lt;&lt; (8*sizeof(int)-2))</code>.
This is the same as <code>2^(8*sizeof(int)-2) - 1 + 2^(8*sizeof(int)-2)</code>.

If <code>sizeof(int) = 4 =&gt; 2^(8*4-2) - 1 + 2^(8*4-2) = 2^30 - 1 + 20^30 = (2^32)/2 - 1 [max signed int of 4 bytes]</code>.

You can't use <code>2*(1 &lt;&lt; (8*sizeof(int)-2)) - 1</code> because it will overflow, but <code>(1 &lt;&lt; (8*sizeof(int)-2)) - 1 + (1 &lt;&lt; (8*sizeof(int)-2))</code> works.

blocks|key|1388810|text|#include+<iostrema>

int+main(){
++++int32_t+maxSigned+=+-1U+>>+1;
++++cout+<<+maxSigned+<<+'\n';
++++return+0;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1388811|它可能依赖于架构，但至少在我的设置中是有效的。|unstyled|1388812|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>#include &lt;iostrema&gt;

int main(){
 int32_t maxSigned = -1U &gt;&gt; 1;
 cout &lt;&lt; maxSigned &lt;&lt; '\n';
 return 0;
}
</code></pre>

It might be architecture dependent but it does work at least in my setup.

Is there any code to find the maximum value of integer (accordingly to the compiler) in C/C++ like <code>Integer.MaxValue</code> function in java?

maximum value of int

像java中的Integer.MaxValue函数一样，C/C++中有没有代码可以找到整数的最大值(对应于编译器)？

问int的最大值
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