将第一个乘以十的第二个位数的幂，然后将另一个相加.

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 的转换(很慢)和 1 个字符串连接(不快)和 1 个 str 到 int 的转换(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，而无需接触主存储器来获取/设置子步骤的数据，这肯定比字符串版本的延迟更小.如果编译器设计将 3-4 个字符串存储在 sse 寄存器中，那么两者都会竞争性能.因为当一个人忙于计算power"函数时，另一个人忙于从 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来构建另一个寄存器.Power(10,x) 函数可以换成 10*10*10.... x 次，所以纯数学版本再次变得更快.

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 imo ***的.

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