PHP不使用内置函数实现字符串转整型的方法示例

$num = '345432123';

 //（一）
$num = (int)$num;
//输出
//int(345432123)

//（二）
$num = intval($num);
//输出
//int(345432123)

function convertInt($strInt = ''){ 
 $len = strlen($strInt); 
 $int = 0;

 for($i=0;$i<$len;$i++){
  $int = 10;   
  $num = $strInt{$i} - '0';   
  $int += $num;  
 }

 return $int;  
}

 $num = '345432123'; 
 var_dump(convertInt($num)); //输出 int(345432123)

#include <stdio.h>
#include <limits.h>
#include <string.h>

/ Convert a string into a long long. Returns 1 if the string could be parsed
  into a (non-overflowing) long long, 0 otherwise. The value will be set to
  the parsed value when appropriate. /
int string2ll(const char s, size_t slen, long long value) {
 const char p = s;
 size_t plen = 0;
 int negative = 0;
 unsigned long long v;

 if (plen == slen)
  return 0;

 / Special case: first and only digit is 0. /
 if (slen == 1 && p[0] == '0') {
  if (value != NULL) value = 0;
  return 1;
 }

 if (p[0] == '-') {
  negative = 1;
  p++; plen++;

  / Abort on only a negative sign. /
  if (plen == slen)
   return 0;
 }

 / First digit should be 1-9, otherwise the string should just be 0. /
 if (p[0] >= '1' && p[0] <= '9') {
  v = p[0]-'0';
  p++; plen++;
 } else if (p[0] == '0' && slen == 1) {
  value = 0;
  return 1;
 } else {
  return 0;
 }

 while (plen < slen && p[0] >= '0' && p[0] <= '9') {
  if (v > (ULLONG_MAX / 10)) / Overflow. /
   return 0;
  v = 10;

  if (v > (ULLONG_MAX - (p[0]-'0'))) / Overflow. /
   return 0;
  v += p[0]-'0';

  p++; plen++;
 }

 / Return if not all bytes were used. /
 if (plen < slen)
  return 0;

 if (negative) {
  if (v > ((unsigned long long)(-(LLONG_MIN+1))+1)) / Overflow. /
   return 0;
  if (value != NULL) value = -v;
 } else {
  if (v > LLONG_MAX) / Overflow. /
   return 0;
  if (value != NULL) value = v;
 }
 return 1;
}

//-------- 执行 ---------
int main(){
 long long num;
 string2ll("345432123",strlen("345432123"),&num);
 printf("%d\n",num); //输出 345432123
 retunr 0;
}