如何在C中定位字符串数组并将其转换为整数？

Question

我有一个要解决的问题。我有一个字符串和整数数组，我只想将整数（163）转换为 C 中的实际整数。

我已设法找到我想要的数字 (163) 数组位置，但我不确定如何将它们转换为数字。我曾尝试使用 strtol、atoi 和 strtoumax，但没有成功。 我在下面添加了我的代码。

char busy[30] = {"this is; it was; 163; 234;;"};

int tag = 0;
int location = 0;
for (int i = sizeof(busy); i > 0; i--)  {
   printf("%c\n", busy[i]); 
   if (busy[i] == ';')  {
        tag = tag+1;
        printf("tag is %i \n", tag);    
   }
   if (tag == 4) {
       printf("for loop is %i\n", i);
       location = i;
       break;
   }
}
location = location+1;
int loca_saved = 0;
int loca_finish = 0;
int tag1 = 0;
while (busy[location] != ';') {
    if (busy[location] == ' ' && busy[location - 1] == ';')  {
    //printf("not printing whitespace between semicolon and characters\n");
    location++;
    }
    else    { 
        if (tag1 == 0)    {
            tag1 = tag1+1;
            loca_saved = location; //this is to tell me the array location for the first char      
        }
        if (busy[location + 1] == ';')  {
            loca_finish = location; //this is to tell me the array location for the last char
        }
    
    putchar(busy[location]); //this is to print my desired characters(163)
    location++;
    }

}

Answer 1

这完全取决于您要解析的确切语法。请注意，例如，负整数以- 字符开头，并且必须有一个最大整数值。由于您的工作非常“低级”，您可能希望在读取字符串时执行转换为整数，例如：

#include <ctype.h>
#include <stdio.h>

int main(void)
{
    char busy[30] = {"this is; it was; 163; 234;;"};

    for (char *p = busy; *p; ++p) // Loop over the string
        if (isdigit(*p)) // Found first digit of an unsigned integer
        {
            unsigned n = (unsigned) (*p - '0'); // Store the value of the first digit in 'n'

            // While reading the number, shift the digits to the appropriate positions
            while (isdigit(*++p))
                n = n * 10U + (unsigned) (*p - '0');

            printf("%u\n", n); // We have finished parsing the integer, print it
        }
}

使用atoi 也可以，但它需要对每个整数进行两次传递：一次在atoi 本身中，另一次在您正在解析的字符串中移动。

strtol 函数允许您将指针传递给您的指针，该指针设置在读取整数之后 的一个位置，从而避免了该问题。您可以使用它来推进指针。在此代码示例中，指针要么手动推进，要么通过strtol：

#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    char busy[30] = {"this is; it was; 163; 234;;"};

    for (char *p = busy; *p; )
        if (isdigit(*p))
            printf("%ld\n", strtol(p, &p, 10));
        else
            ++p;
}

另一种选择，不使用isdigit：

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    char busy[30] = {"this is; it was; 163; 234;;"};

    for (char *p = busy; *p; )
    {
        char *end;
        long n = strtol(p, &end, 10);
        if (p != end) // Pointer was advanced, so an integer was read
        {
            printf("%ld\n", n);
            p = end;
        }
        else // Pointer was not advanced, do so manually
            ++p;
    }
}

Answer 2

strspn 和 strcspn 可用于解析字符串。
可以使用strtol 代替sscanf 来获取号码。

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

int main ( void) {
    char busy[] = "this is; it was; 163; 234;";
    char *dlm = "0123456789";
    char *parse = busy;
    int number = 0;

    while ( *parse) {
        parse += strcspn ( parse, dlm);//count to next delimiter
        if ( 1 == sscanf ( parse, "%d", &number)) {
            printf ( "%d\n", number);
        }
        parse += strspn ( parse, dlm);//skip delimeters
    }

    return 0;
}

解析也可以去掉分号。
有些字段没有整数。

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

int main ( void) {
    char busy[] = "this is; it was; 163; 234;";
    char *dlm = ";";
    char *parse = busy;
    int number = 0;

    while ( *parse) {
        if ( 1 == sscanf ( parse, "%d", &number)) {
            printf ( "%d\n", number);
        }
        else {
            printf ( "could not parse integer\n");
        }
        parse += strcspn ( parse, dlm);//count to next delimiter
        parse += strspn ( parse, dlm);//skip delimeters
        //the above line will skip all consecutive delimiters
        //to process each delimiter use the line below
        //++parse;//skip one delimiter
    }

    return 0;
}

Answer 3

由于您有一个可修改的字符串，您可以使用strtok 将其分隔为基于; 分隔符的子字符串。如果子字符串是数字，则此代码会执行此操作。

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

int main (void)
{
  char str[30] = {"this is; it was; 163; 234;;"};
  
  for(char* p=strtok(str, ";"); p!=NULL; p=strtok(NULL,str))
  {
    char* endptr;
    int i = strtol(p,&endptr,10);
    if(endptr != p)
    {
      printf("%d\n", i);
    }
  }
}

输出：

163
234

strtol 将“endptr”参数设置为指向字符串的开头以防万一失败，因此我们可以使用它来确定子字符串是否为数字。