sscanf 用法 - 如何验证完成的扫描与中止的扫描

问题描述

我的数据库提供了一个文本文件，其中包含打开和关闭 " 来分隔公式.这组公式非常有限，一旦确定就很容易实现.我尝试使用 scanf 来获取参数，我想使用分隔符 " 来提供 scanf 失败的机制.

在下面的示例中，最后一个分隔符被忽略，并且找不到分隔符的信息丢失.我如何控制 sscanf 是否能够匹配整个字符串?

#include <stdio.h>#include <string.h>无符号 printIdentity(const char * 公式){无符号 e = 0，找到 = 0；双a，b；printf("-------------
");printf("调查: %s
", 公式);if( ( ( 2 == sscanf_s( 公式, " " X * %lf %lf " ", &a, &b, sizeof( double ), sizeof( double ) ) ) ){printf("匹配: X * %lf + %lf
", a, b);++找到；}if( ( ( 1 == sscanf_s( 公式, " " X * %lf " ", &a, sizeof( double ) ) ) ){printf("比赛: X * %lf
", a);++找到；}如果(找到！= 1){e += 1;printf("错误: %u 公式类型
", 找到);}printf("-------------
");返回 e;}无符号主(无效){无符号 e = 0;e += printIdentity(" "X*3.1"");e += printIdentity(" "X*3.2-4.2"");e += printIdentity(" "X*3.3+4.3"");if( 0 != e ){ printf( "错误: %2u
", e );}else{ printf("全部通过
", e);}返回 e;}

解决方案

使用格式说明符%n获取处理结束的位置并与输入字符串的长度进行比较，其中n格式说明符定义为(从第 7.19.6.2 节 C99 标准的 fscanf 函数):

不消耗任何输入.对应的参数应该是一个指向要写入的有符号整数 从中读取的字符数到目前为止，通过调用 fscanf 函数得到的输入流.执行一个%n 指令不会增加在fscanf 函数的执行完成.没有参数被转换，但一个被消耗了.如果转换规范包括分配抑制字符或字段宽度，行为未定义.

例如:

#include <string.h>#include <stdlib.h>#include <stdio.h>主函数(){const char* good_input = ""20 21"";const char* bad_input = ""14 12";诠释一个[2]；位置；if (sscanf(good_input, " "%d %d"%n", &a[0], &a[1], &pos) == 2 &&pos == strlen(good_input)){printf("good_input: %d %d
", a[0], a[1]);}if (sscanf(bad_input, " "%d %d"%n", &a[0], &a[1], &pos) == 2 &&pos == strlen(bad_input)){printf("bad_input: %d %d
", a[0], a[1]);}}

输出:

好输入:20 21

http://codepad.org/6G4lLXWg 上的在线演示.

My database provides a textfile with opening and closing " to delimiter formulas.The set of formulas is very limited and will be easy to implement once identified.I try to use scanf to get the parameters and i want to use the delimiter " to provide a mechanism for scanf to fail.

In the below example the last delimiter is ignored and the information that the delimiter was not found is lost. How can i control if sscanf was able to match the whole string?

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

unsigned printIdentity(const char * formula){
    unsigned e = 0, found = 0;
    double a, b;
    printf("-------------
");
    printf("INVESTIGATING: %s
", formula);
    if( ( 2 == sscanf_s( formula, " " X * %lf %lf " ", &a, &b, sizeof( double ), sizeof( double ) ) ) ){
        printf("MATCH: X * %lf + %lf
", a, b);
        ++found;
    }
    if( ( 1 == sscanf_s( formula, " " X * %lf " ", &a, sizeof( double ) ) ) ){
        printf("MATCH: X * %lf
", a);
        ++found;
    }
    if( found != 1){
        e += 1;
        printf("ERROR: %u formula types
", found);
    }
    printf("-------------
");
    return e;
}

unsigned main( void )
{
    unsigned e = 0;

    e += printIdentity("     "X*3.1"");
    e += printIdentity("     "X*3.2-4.2"");
    e += printIdentity("     "X*3.3+4.3"");

    if( 0 != e ){ printf( "ERRORS: %2u
", e ); }
    else{ printf( "all pass
", e ); }
    return e;
}

解决方案

Use the format specifier %n to obtain the position where processing ended and compare with the length of the input string, where the n format specifier is defined as (from section 7.19.6.2 The fscanf function of the C99 standard):

For example:

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

int main()
{
    const char* good_input = ""20 21"";
    const char* bad_input = ""14 12";
    int a[2];
    int pos;

    if (sscanf(good_input, " "%d %d"%n", &a[0], &a[1], &pos) == 2 &&
        pos == strlen(good_input))
    {
        printf("good_input: %d %d
", a[0], a[1]);
    }

    if (sscanf(bad_input, " "%d %d"%n", &a[0], &a[1], &pos) == 2 &&
        pos == strlen(bad_input))
    {
        printf("bad_input: %d %d
", a[0], a[1]);
    }
}

Output:

good_input: 20 21

Online demo at http://codepad.org/6G4lLXWg .

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