C语言 无需访问库即可将整数转换为字符串
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原文地址: http://stackoverflow.com/questions/3982320/
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Convert integer to string without access to libraries
提问by Nick Sinas
I recently read a sample job interview question:
我最近阅读了一个示例工作面试问题：
Write a function to convert an integer to a string. Assume you do not have access to library functions i.e., itoa(), etc...
编写一个将整数转换为字符串的函数。假设您无权访问库函数，即 itoa() 等...
How would you go about this?
你会怎么做？
采纳答案by John Boker
fast stab at it: (edited to handle negative numbers)
快速刺杀它：（编辑以处理负数）
int n = INT_MIN;
char buffer[50];
int i = 0;
bool isNeg = n<0;
unsigned int n1 = isNeg ? -n : n;
while(n1!=0)
{
buffer[i++] = n1%10+'0';
n1=n1/10;
}
if(isNeg)
buffer[i++] = '-';
buffer[i] = 'char* itoa(int val, int base){
static char buf[32] = {0};
int i = 30;
for(; val && i ; --i, val /= base)
buf[i] = "0123456789abcdef"[val % base];
return &buf[i+1];
}
';
for(int t = 0; t < i/2; t++)
{
buffer[t] ^= buffer[i-t-1];
buffer[i-t-1] ^= buffer[t];
buffer[t] ^= buffer[i-t-1];
}
if(n == 0)
{
buffer[0] = '0';
buffer[1] = 'unsigned countDigits(long long x)
{
int i = 1;
while ((x /= 10) && ++i);
return i;
}
unsigned getNumDigits(long long x)
{
x < 0 ? x = -x : 0;
return
x < 10 ? 1 :
x < 100 ? 2 :
x < 1000 ? 3 :
x < 10000 ? 4 :
x < 100000 ? 5 :
x < 1000000 ? 6 :
x < 10000000 ? 7 :
x < 100000000 ? 8 :
x < 1000000000 ? 9 :
x < 10000000000 ? 10 : countDigits(x);
}
#define tochar(x) '0' + x
void tostr(char* dest, long long x)
{
unsigned i = getNumDigits(x);
char negative = x < 0;
if (negative && (*dest = '-') & (x = -x) & i++);
*(dest + i) = 0;
while ((i > negative) && (*(dest + (--i)) = tochar(((x) % 10))) | (x /= 10));
}
';
}
printf(buffer);
回答by dnbwise
The algorithm is easy to see in English.
该算法在英文中很容易看到。
Given an integer, e.g. 123
给定一个整数，例如 123
divide by 10 => 123/10. Yielding, result = 12 and remainder = 3
add 30h to 3 and push on stack (adding 30h will convert 3 to ASCII representation)
repeat step 1 until result < 10
add 30h to result and store on stack
the stack contains the number in order of | 1 | 2 | 3 | ...
除以 10 => 123/10。产量，结果 = 12，余数 = 3
将 30h 添加到 3 并压入堆栈（添加 30h 会将 3 转换为 ASCII 表示）
重复步骤 1 直到结果 < 10
将 30h 添加到结果并存储在堆栈中
堆栈包含按顺序排列的数字 | 1 | 2 | 3 | ...
回答by Benoit Thiery
A look on the web for itoa implementation will give you good examples. Here is one, avoiding to reverse the string at the end. It relies on a static buffer, so take care if you reuse it for different values.
在网上查看 itoa 实现会给你很好的例子。这是一个，避免在最后反转字符串。它依赖于静态缓冲区，因此如果将其重用于不同的值，请务必小心。
// Return character one past end of character digits.
static char *myitoa_helper(char *dest, int neg_a) {
if (neg_a <= -10) {
dest = myitoa_helper(dest, neg_a / 10);
}
*dest = (char) ('0' - neg_a % 10);
return dest + 1;
}
char *myitoa(char *dest, int a) {
if (a >= 0) {
*myitoa_helper(dest, -a) = '#include "limits.h"
#include "stdio.h"
int main(void) {
const int a[] = {INT_MIN, INT_MIN + 1, -42, -1, 0, 1, 2, 9, 10, 99, 100,
INT_MAX - 1, INT_MAX};
for (unsigned i = 0; i < sizeof a / sizeof a[0]; i++) {
myitoa_test(a[i]);
}
return 0;
}
-2147483648 <-2147483648>
-2147483647 <-2147483647>
-42 <-42>
-1 <-1>
0 <0>
1 <1>
2 <2>
9 <9>
10 <10>
99 <99>
100 <100>
2147483646 <2147483646>
2147483647 <2147483647>
';
} else {
*dest = '-';
*myitoa_helper(dest + 1, a) = ' int num = ...;
char res[MaxDigitCount];
int len = 0;
for(; num > 0; ++len)
{
res[len] = num%10+'0';
num/=10;
}
res[len] = 0; //null-terminating
//now we need to reverse res
for(int i = 0; i < len/2; ++i)
{
char c = res[i]; res[i] = res[len-i-1]; res[len-i-1] = c;
}
';
}
return dest;
}
void myitoa_test(int a) {
char s[100];
memset(s, 'x', sizeof s);
printf("%11d <%s>\n", a, myitoa(s, a));
}
回答by Beyondo
The faster the better?
越快越好吗？
char *pru(unsigned x, char *eob)
{
do { *--eob = x%10; } while (x/=10);
return eob;
}
char *pri(int x, char *eob)
{
eob = fmtu(x<0?-x:x, eob);
if (x<0) *--eob='-';
return eob;
}
If you want to debug, You can split the conditions (instructions) into
lines of code inside the while scopes {}.
如果要调试，可以将条件（指令）拆分
为 while 范围内的代码行{}。
回答by chux - Reinstate Monica
Convert integer to string without access to libraries
无需访问库即可将整数转换为字符串
Convert the least significant digit to a character first and then proceed to more significant digits.
先将最低有效数字转换为字符，然后再转换为更高有效数字。
Normally I'd shift the resulting stringinto place, yet recursion allows skipping that step with some tight code.
通常我会将结果字符串移到位，但递归允许使用一些紧凑的代码跳过该步骤。
Using neg_ain myitoa_helper()avoids undefined behavior with INT_MIN.
使用neg_ainmyitoa_helper()避免了未定义的行为INT_MIN。
char *SignedIntToStr(char *Dest, signed int Number, register unsigned char Base) {
if (Base < 2 || Base > 36) {
return (char *)0;
}
register unsigned char Digits = 1;
register unsigned int CurrentPlaceValue = 1;
for (register unsigned int T = Number/Base; T; T /= Base) {
CurrentPlaceValue *= Base;
Digits++;
}
if (!Dest) {
Dest = malloc(Digits+(Number < 0)+1);
}
char *const RDest = Dest;
if (Number < 0) {
Number = -Number;
*Dest = '-';
Dest++;
}
for (register unsigned char i = 0; i < Digits; i++) {
register unsigned char Digit = (Number/CurrentPlaceValue);
Dest[i] = (Digit < 10? '0' : 87)+Digit;
Number %= CurrentPlaceValue;
CurrentPlaceValue /= Base;
}
Dest[Digits] = '##代码##';
return RDest;
}
#include <stdio.h>
int main(int argc, char *argv[]) {
char String[32];
puts(SignedIntToStr(String, -100, 16));
return 0;
}
Test code & output
测试代码和输出
##代码##回答by eyalm
An implementation of itoa()function seems like an easy task but actually you have to take care of many aspects that are related on your exact needs. I guess that in the interview you are expected to give some details about your way to the solution rather than copying a solution that can be found in Google (http://en.wikipedia.org/wiki/Itoa)
itoa()函数的实现似乎是一项简单的任务，但实际上您必须处理与您的确切需求相关的许多方面。我想在面试中你应该提供一些关于解决方案的细节，而不是复制可以在谷歌找到的解决方案（http://en.wikipedia.org/wiki/Itoa）
Here are some questions you may want to ask yourself or the interviewer:
以下是您可能想问自己或面试官的一些问题：
- Where should the string be located (malloced? passed by the user? static variable?)
- Should I support signed numbers?
- Should i support floating point?
- Should I support other bases rather then 10?
- Do we need any input checking?
- Is the output string limited in legth?
- 字符串应该位于哪里（malloced？由用户传递？静态变量？）
- 我应该支持签名数字吗？
- 我应该支持浮点数吗？
- 我应该支持其他基础而不是 10 个吗？
- 我们需要任何输入检查吗？
- 输出字符串是否有长度限制？
And so on.
等等。
回答by nategoose
I would keep in mind that all of the digit characters are in increasing order within the ASCII character set and do not have other characters between them.
我会记住，所有数字字符在 ASCII 字符集中都是按递增顺序排列的，并且它们之间没有其他字符。
I would also use the /and the%operators repeatedly.
我也会反复使用/和%运算符。
How I would go about getting the memory for the string would depend on information you have not given.
我将如何获取字符串的内存取决于您未提供的信息。
回答by Armen Tsirunyan
Assuming it is in decimal, then like this:
假设它是十进制的，那么像这样：
##代码##回答by R.. GitHub STOP HELPING ICE
Here's a simple approach, but I suspect if you turn this in as-is without understanding and paraphrasing it, your teacher will know you just copied off the net:
这是一个简单的方法，但我怀疑如果你在没有理解和解释的情况下按原样转换它，你的老师会知道你只是从网上复制的：
##代码##Various improvements are possible, especially if you want to efficiently support larger-than-word integer sizes up to intmax_t. I'll leave it to you to figure out the way these functions are intended to be called.
各种改进都是可能的，特别是如果您想有效地支持高达intmax_t. 我会让你去弄清楚这些函数的调用方式。
回答by No Body
I came across this question so I decided to drop by the code I usually use for this:
我遇到了这个问题，所以我决定放弃我通常用于此的代码：
##代码##This will automatically allocate memory if NULL is passed into Dest. Otherwise it will write to Dest.
如果将 NULL 传递给 Dest，这将自动分配内存。否则它将写入 Dest。
