判断输入字符串是否为浮点数
写一个Bool型的函数，要求自己实现
[解决办法]
浮点数可以由以下的正则表达式确定：white_space_opt [[:space:]]*digit_sequence [[:digit:]]+fractional_constant {digit_sequence}\.?|\.{digit_sequence}|{digit_sequence}\.{digit_sequence}exponent_part [eE][+-]?{digit_sequence}floating_constant {fractional_constant}{exponent_part}*accepted_input {white_space_opt}[+-]?{floating_constant}{white_space_opt}---------------------------------------------如果不接受 1 2 3 这样的数为浮点数， 则fractional_constant应该是 {digit_sequence}\.|.....如果不接受 浮点数两端加的空格,则 accepted_input 应该是 [+-]?{floating_constant}---------------------------------------------你可以用你稀饭的正则表达式库，如posix regex , pcre , boost::regex 啥的做这件事，也可以用其对应的 DFA 做判断， 比如这样#include <stdio.h>int is_floating_constant( const char* input ){static const int yy_accept[21] = { 0, 0, 0, 3, 2, 2, 2, 2, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 0 } ;static const int yy_ec[256] = { 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 4, 1, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 } ;static const int yy_meta[7] = { 0, 1, 2, 1, 1, 3, 1 } ;static const int yy_base[24] = { 0, 0, 0, 27, 28, 5, 7, 0, 11, 0, 0, 0, 0, 16, 0, 21, 20, 0, 0, 19, 28, 17, 17, 11 } ;static const int yy_def[24] = { 0, 20, 1, 20, 20, 20, 20, 21, 20, 5, 6, 21, 8, 20, 22, 13, 20, 15, 23, 15, 0, 20, 20, 20 } ;static const int yy_nxt[35] = { 0, 4, 5, 6, 7, 8, 4, 9, 10, 11, 12, 11, 12, 14, 19, 15, 12, 16, 14, 14, 13, 13, 16, 18, 19, 19, 17, 20, 3, 20, 20, 20, 20, 20, 20 } ;static const int yy_chk[35] = { 0, 1, 1, 1, 1, 1, 1, 5, 5, 5, 5, 6, 6, 8, 23, 8, 8, 8, 13, 22, 21, 13, 13, 16, 19, 16, 15, 3, 20, 20, 20, 20, 20, 20, 20 } ; enum { last_dfa = 19 , jambase = 28 }; int curr = 1 , c ; do { if( 0 == *input ) break; c = yy_ec[*(unsigned char*)input]; if( 1 == yy_accept[curr] && 0 == input[1] ) return 1; while( yy_chk[ yy_base[curr] + c ] != curr ) if( (curr=yy_def[curr] ) >= last_dfa + 2 ) c = yy_meta[c]; curr = yy_nxt[ yy_base[curr] + c ]; ++input; } while( yy_base[curr] != jambase ); return 0;}int main(){ char input[80+1] = ""; while( fgets( input , 80 , stdin ) ) { if( is_floating_constant( input ) ) printf( "YES\n" ); else printf( "NO\n" ); } return 0;}