[转]sscanf高级用法总结

    技术2022-05-11  66

    大家都知道sscanf是一个很好用的函数,利用它可以从字符串中取出整数、浮点数和字符串等等。它的使用方法简单,特别对于整数和浮点数来说。但新手可能并不知道处理字符串时的一些高级用法,这里做个简要说明吧。1. 常见用法。char str[512] = {0};sscanf("123456 ", "%s", str);printf("str=%s/n", str);2. 取指定长度的字符串。如在下例中,取最大长度为4字节的字符串。sscanf("123456 ", "%4s", str);printf("str=%s/n", str);3. 取到指定字符为止的字符串。如在下例中,取遇到空格为止字符串。sscanf("123456 abcdedf", "%[^ ]", str);printf("str=%s/n", str);4. 取仅包含指定字符集的字符串。如在下例中,取仅包含1到9和小写字母的字符串。sscanf("123456abcdedfBCDEF", "%[1-9a-z]", str);printf("str=%s/n", str);5. 取到指定字符集为止的字符串。如在下例中,取遇到大写字母为止的字符串。sscanf("123456abcdedfBCDEF", "%[^A-Z]", str);printf("str=%s/n", str);

     源代码一如下:#include <stdio.h>#include <stdlib.h>

    char *tokenstring = "12:34:56-7890";char a1[3], a2[3], a3[3];int i1, i2;

    void main(void){   clrscr();   sscanf(tokenstring,  "%2s:%2s:%2s---",  a1, a2, a3, &i1, &i2);   printf("%s/n%s/n%s/n%d/n%d/n/n", a1, a2, a3, i1, i2);   getch();}

    源代码二如下:#include <stdio.h>#include <stdlib.h>

    char *tokenstring = "12:34:56-7890";char a1[3], a2[3], a3[3];int i1, i2;

    void main(void){   clrscr();   sscanf(tokenstring,  "%2s%1s%2s%1s%2s%1s--",  a1,  &a, a2, &a3, a3, &a, &i1, &i2);   printf("%s/n%s/n%s/n%d/n%d/n/n", a1, a2, a3, i1, i2);   getch();}

    源代码三如下:#include <stdio.h>#include <stdlib.h>

    char *tokenstring = "12:34:56-7890";char a1[3], a2[3], a3[3], a4[3], a5[3];int i1, i2;

    void main(void){   char a;

       clrscr();   sscanf(tokenstring,  "%2s%1s%2s%1s%2s%1s%2s%2s",  a1,  &a, a2, &a3, a3, &a, a4, a5);   i1 =atoi(a4);   i2 =atoi(a5);

       printf("%s/n%s/n%s/n%d/n%d/n/n", a1, a2, a3, i1, i2);   getch();}

    方法四如下(以实例说明,原理相同):/* The following sample illustrates the use of brackets and the   caret (^) with sscanf().   Compile options needed: none*/

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

    char *tokenstring = "first,25.5,second,15";int result, i;double fp;char o[10], f[10], s[10], t[10];

    void main(){   result = sscanf(tokenstring, "%[^','],%[^','],%[^','],%s", o, s, t, f);   fp = atof(s);   i  = atoi(f);   printf("%s/n %lf/n %s/n %d/n", o, fp, t, i);}


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