📅  最后修改于: 2020-09-25 08:59:53             🧑  作者: Mango
C++中的strtol() 函数将字符串的内容解释为指定基数的整数,并将其值返回为long int。此函数还将设置一个指针,以指向该字符串的最后一个有效字符之后的第一个字符 (如果存在),否则将其设置为null。
For base 10 and the string "12abc":
Valid numeric part -> 12
First character after valid numeric part -> a
long int strtol(const char* str, char** end, int base);
strtol() 函数将字符串,指向字符的指针和整数值-base作为其参数,将字符串的内容解释为给定base的整数,并返回一个long int值。
此函数在
strtol() 函数返回:
#include
#include
using namespace std;
int main()
{
int base = 10;
char str[] = "27ab_1x";
char *end;
long int num;
num = strtol(str, &end, base);
cout << "Number in String = " << str << endl;
cout << "Number in Long Int = " << num << endl;
cout << "End String = " << end << endl << endl;
// the pointer to invalid characters can be null
strcpy(str, "27");
cout << "Number in String = " << str << endl;
num = strtol(str, &end, base);
cout << "Number in Long Int = " << num << endl;
if (*end) {
cout << end;
} else {
cout << "Null pointer";
}
return 0;
}
运行该程序时,输出为:
Number in String = 27ab_1x
Number in Long Int = 27
End String = ab_1x
Number in String = 27
Number in Long Int = 27
Null pointer
strtol() 函数的有效整数值包括:
参数base的有效值为{0,2,3,...,35,36}。以2为底的一组有效数字是{0,1},以3为底的一组有效数字是{0,1,2},依此类推。对于从11到36的基数,有效数字包括字母。底数11的有效数字集为{0,1,…,9,A,a},底数12的有效数字为{0,1,…,9,A,a,B,b},依此类推。
注意:请务必记住,一个基数的有效字符可能会以另一基数的无效字符串 ,如下例所示。
#include
#include
#include
using namespace std;
int main()
{
char *end;
cout << "128bz" << " to Long Int with base-5 = " << strtol("128bxz", &end, 5) << endl;
cout << "End String = " << end << endl << endl;
cout << "128bz" << " to Long Int with base-12 = " << strtol("128bxz", &end, 12) << endl;
cout << "End String = " << end << endl << endl;
cout << "128bz" << " to Long Int with base-36 = " << strtol("128bxz", &end, 36) << endl;
cout << "End String = " << end << endl << endl;
return 0;
}
运行该程序时,输出为:
128bz to Long Int with base-5 = 7
End String = 8bxz
128bz to Long Int with base-12 = 2123
End String = xz
128bz to Long Int with base-36 = 64214135
End String =
直到主非空白字符被找到与strtol() 函数将忽略所有的前导空白字符 。
通常,strtol() 函数的有效整数参数具有以下形式:
[whitespace] [- | +] [0 | 0x] [alphanumeric characters]
然后,从这个字符开始,它需要尽可能多的字符来形成有效的整数表示形式并将其转换为long int值。最后一个有效字符之后的字符串剩余部分将被忽略,并且对结果没有任何影响。
#include
#include
using namespace std;
int main()
{
char *end;
cout << " 25axbz" << " to Long Int with base-11 = " << strtol(" 25axbz", &end, 11) << endl;
cout << "End String = " << end << endl << endl;
cout << " 110bcd" << " to Long Int with base-2 = " << strtol(" 110bcd", &end, 2) << endl;
cout << "End String = " << end << endl << endl;
cout << "ax110.97" << " to Long Int with base-10 = " << strtol("ax110.97", &end, 10) << endl;
cout << "End String = " << end << endl << endl;
return 0;
}
运行该程序时,输出为:
25axbz to Long Int with base-11 = 307
End String = xbz
110bcd to Long Int with base-2 = 6
End String = bcd
ax110.97 to Long Int with base-10 = 0
End String = ax110.97
如果基数为``0'',则通过查看字符串格式自动确定数字基数。如果前缀为0,则基数为八进制(8)。如果前缀为0x或0X,则基数为十六进制(16),否则基数为十进制(10)。
#include
#include
using namespace std;
int main()
{
char *end;
/* octal base */
cout << "0128ai" << " to Long Int with base-0 = " << strtol("0128ai", &end, 0) << endl;
cout << "End String = " << end << endl << endl;
/* hexadecimal base */
cout << "0x15axzz" << " to Long Int with base-0 = " << strtol("0x15axzz", &end, 0) << endl;
cout << "End String = " << end << endl << endl;
/* decimal base */
cout << "23dfl" << " to Long Int with base-0 = " << strtol("23dfl", &end, 0) << endl;
cout << "End String = " << end << endl << endl;
return 0;
}
运行该程序时,输出为:
0128ai to Long Int with base-0 = 10
End String = 8ai
0x15axzz to Long Int with base-0 = 346
End String = xzz
23dfl to Long Int with base-0 = 23
End String = dfl