如何在 C++ 中附加一个整数(带一个整数) [英] How To Append an Integer (with an Integer) in C++

问题描述

我想知道是否有人可以告诉我如何在 C++ 中附加一个整数(和另一个整数).基本上，如果我有一个值为 67 的 int，我如何将它附加到数字 4 以便整数现在是 674?提前致谢！

I was wondering if anyone could tell me how to append an integer (with another integer) in C++. Basically, if I have an int with this the value 67, how would I append it with the number 4 so the integer is now 674? Thanks in advance!

推荐答案

先乘以 10 的秒数的幂，然后再加上另一个.

Multiply first by ten to the power of digit number of second and add the other .

示例:63 和 5

63*10=630
630+5 =635

示例:75 和 34

75*100=7500
7500+34=7534
int i1=75;
int i2=34;
int dn=ceil(log10(i2+0.001));     //0.001 is for exact 10, exact 100, ...
int i3=i1*ceil(pow(10,dn)); <---- because pow would give 99.999999(for some optimization modes)
i3+=i2;

字符串版本需要 2 个 int 到 str 转换(很慢)和 1 个字符串连接(不快)和 1 个 str 到 int 转换(很慢).上转换需要 2 次加法、1 次对数、2 次上限、1 次幂、1 次乘法，所有这些都可以在 cpu 中完成，而无需接触主内存来获取/设置子步骤的数据，这肯定比字符串版本的延迟更少.如果编译器设计将 3-4 个字符串存储在 sse 寄存器中，那么两者都会争夺性能.因为当一个人忙于计算功率"函数时，另一个人会忙于从 sse 中提取字符串并将其一个一个地放入必要的寄存器并通过开始加法和乘法在另一个寄存器上构造.Power(10,x) 函数可以换取 10*10*10.. x 倍，所以纯数学版本又变得更快了.

String version needs 2 int to str conversion (which is slow) and 1 string concatenation (which is not fast) and 1 str to int conversion (which is slow). Upper conversion needs 2 additions, 1 logarithm, 2 ceilings, 1 power, 1 multiplication all of which could be done in cpu without touching main memory to get/set data for sub steps that is surely less latency then string versions. If 3-4 character strings are stored in sse registers by compiler design, then both would compete for performance. Because while one would be busy computing "power" function, other would be busy extracting string from sse and putting it necessary registers one by one and constructing on another register by starting additions and multiplications. Power(10,x) function can be traded for 10*10*10.... x times so pure math version becomes faster again.

如果您需要可读性，那么 eq- 的答案是最好的.

If it is readability you need, eq- 's answer is best imo.

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