回答編集履歴
2
追記
test
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@@ -20,9 +20,45 @@
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**追記1**
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enumと構造体を使ってオペレータか数字の配列を作る
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```C
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enum {
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OP_ADD,
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OP_SUB,
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OP_MUL,
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OP_DIV,
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NUMEBR,
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}KIND;
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struct {
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KIND kind;
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int value;
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}ELEMENT;
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ELEMENT elements[30];
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```
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25
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**追記**
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**追記2**
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()を総当たりで挿入する(オペレータは1文字、スペースは含まない前提)
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64
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1
追記
test
CHANGED
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@@ -17,3 +17,123 @@
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3が難しかったらとりあえず式を文字列として見て、数字の前に(、それ以降の数字の後ろに)を挿入した文字列を順次作成していけば、あとは一般的な式の評価を行えばいいと思います。
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**追記**
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()を総当たりで挿入する(オペレータは1文字、スペースは含まない前提)
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30
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```C
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static int evaluateExpression(const char* exp)
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{
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printf("eval: %s\n", exp);
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return 0; // none
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}
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45
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static const char* passNumber(const char* s)
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{
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while (*s && isdigit(*s)) s++;
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return s;
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}
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int main()
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{
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const char* exp = "18-7*3-1+2=6";
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64
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const char* openPara = exp;
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const char* closePara;
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68
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char exp2[100];
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while (1) {
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closePara = passNumber(openPara);
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while (*closePara != '=') {
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closePara = passNumber(closePara + 1); // +1 : pass operator
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85
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char* d = exp2;
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for (const char* s = exp; *s; s++) {
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if (s == openPara) *d++ = '(';
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if (s == closePara) *d++ = ')';
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*d++ = *s;
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}
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100
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*d = '\0';
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102
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103
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104
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105
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if (evaluateExpression(exp2)) {
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106
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107
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printf("anser is : %s\n", exp2);
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108
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109
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goto EXIT:
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110
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111
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}
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112
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113
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}
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114
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115
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116
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117
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openPara = passNumber(openPara);
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118
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119
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if (*openPara == '=') {
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120
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121
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break;
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122
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123
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}
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124
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125
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openPara++; // pass operator
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126
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127
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}
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128
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129
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EXIT:
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130
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131
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132
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133
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return 0;
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134
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135
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}
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136
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137
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```
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138
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139
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あとは[このへんを参考に](http://blog.goo.ne.jp/masaki_goo_2006/e/754f937f81f8b3ac389b9265c52fd98d)式を評価してみる
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