回答編集履歴
4
strtol使用
answer
CHANGED
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@@ -7,14 +7,6 @@
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7
7
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8
8
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regex_t preg[3];
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9
9
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10
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-
int getValue(char *f, int so, int eo) {
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11
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-
char backup = f[eo];
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12
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-
f[eo] = '\0';
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13
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-
int v = atoi(&f[so]);
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14
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-
f[eo] = backup;
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15
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-
return v;
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16
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-
}
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17
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-
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18
10
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void replace(char *f, int so, int eo, int v) {
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19
11
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sprintf(&f[so], "%d", v);
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20
12
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strcat(f, &f[eo]);
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@@ -30,15 +22,15 @@
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30
22
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}
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31
23
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// *
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32
24
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while(!regexec(&preg[1], f, 3, match, 0)) {
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33
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-
int v1 =
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25
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+
int v1 = (int)strtol(&f[match[1].rm_so], NULL, 10);
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34
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-
int v2 =
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26
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+
int v2 = (int)strtol(&f[match[2].rm_so], NULL, 10);
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35
27
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int v = v1 * v2;
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36
28
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replace(f, match[0].rm_so, match[0].rm_eo, v);
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37
29
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}
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38
30
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// +
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39
31
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while(!regexec(&preg[2], f, 3, match, 0)) {
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40
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-
int v1 =
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32
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+
int v1 = (int)strtol(&f[match[1].rm_so], NULL, 10);
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41
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-
int v2 =
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33
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+
int v2 = (int)strtol(&f[match[2].rm_so], NULL, 10);
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42
34
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int v = v1 + v2;
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43
35
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replace(f, match[0].rm_so, match[0].rm_eo, v);
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44
36
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}
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@@ -221,7 +213,6 @@
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221
213
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test("1+3*(2+4)");
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222
214
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}
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223
215
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```
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224
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-
実行
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225
216
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```plain
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226
217
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1 + 2
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227
218
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1 2 +
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3
正規表現修正
answer
CHANGED
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@@ -48,7 +48,7 @@
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48
48
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void test(char *formula) {
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49
49
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printf("f=%s\n", formula);
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50
50
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51
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-
regcomp(&preg[0], "\(
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51
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+
regcomp(&preg[0], "\([^(]+\)", REG_EXTENDED|REG_NEWLINE); // ()
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52
52
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regcomp(&preg[1], "([0-9]+)\*([0-9]+)", REG_EXTENDED|REG_NEWLINE); // *
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53
53
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regcomp(&preg[2], "([0-9]+)\+([0-9]+)", REG_EXTENDED|REG_NEWLINE); // +
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54
54
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2
正規表現コード追加
answer
CHANGED
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@@ -1,5 +1,92 @@
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1
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+
正規表現を使えばこうなるでしょう。
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2
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+
```c
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3
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+
#include <stdio.h>
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4
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+
#include <stdlib.h>
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5
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+
#include <string.h>
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6
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+
#include <regex.h>
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7
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+
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8
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+
regex_t preg[3];
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9
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+
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10
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+
int getValue(char *f, int so, int eo) {
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11
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+
char backup = f[eo];
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12
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+
f[eo] = '\0';
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13
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+
int v = atoi(&f[so]);
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14
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+
f[eo] = backup;
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15
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+
return v;
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16
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+
}
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17
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+
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18
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+
void replace(char *f, int so, int eo, int v) {
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19
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+
sprintf(&f[so], "%d", v);
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20
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+
strcat(f, &f[eo]);
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21
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+
}
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22
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+
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23
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+
int calc(char *f) {
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24
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+
regmatch_t match[3];
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25
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+
// ()
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26
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+
while(!regexec(&preg[0], f, 1, match, 0)) {
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27
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+
f[(int)match[0].rm_eo-1] = '\0';
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28
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+
int v = calc(&f[match[0].rm_so+1]);
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29
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+
replace(f, match[0].rm_so, match[0].rm_eo, v);
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30
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+
}
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31
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+
// *
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32
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+
while(!regexec(&preg[1], f, 3, match, 0)) {
|
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33
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+
int v1 = getValue(f, match[1].rm_so, match[1].rm_eo);
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34
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+
int v2 = getValue(f, match[2].rm_so, match[2].rm_eo);
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35
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+
int v = v1 * v2;
|
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36
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+
replace(f, match[0].rm_so, match[0].rm_eo, v);
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37
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+
}
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38
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+
// +
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39
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+
while(!regexec(&preg[2], f, 3, match, 0)) {
|
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40
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+
int v1 = getValue(f, match[1].rm_so, match[1].rm_eo);
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41
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+
int v2 = getValue(f, match[2].rm_so, match[2].rm_eo);
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42
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+
int v = v1 + v2;
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43
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+
replace(f, match[0].rm_so, match[0].rm_eo, v);
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44
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+
}
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45
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+
return atoi(f);
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|
46
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+
}
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47
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+
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48
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+
void test(char *formula) {
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49
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+
printf("f=%s\n", formula);
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50
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+
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51
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+
regcomp(&preg[0], "\(.+\)", REG_EXTENDED|REG_NEWLINE); // ()
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52
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+
regcomp(&preg[1], "([0-9]+)\*([0-9]+)", REG_EXTENDED|REG_NEWLINE); // *
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53
|
+
regcomp(&preg[2], "([0-9]+)\+([0-9]+)", REG_EXTENDED|REG_NEWLINE); // +
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54
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+
|
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55
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+
char *f = malloc(strlen(formula)+1);
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56
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+
strcpy(f, formula);
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57
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+
|
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58
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+
int a = calc(f);
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59
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+
printf("a=%d\n", a);
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60
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+
|
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61
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+
free(f);
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62
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+
|
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63
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+
regfree(&preg[2]);
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64
|
+
regfree(&preg[1]);
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|
65
|
+
regfree(&preg[0]);
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66
|
+
}
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|
67
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+
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68
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+
int main(int argc, char *argv[]) {
|
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69
|
+
test("1+2");
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|
70
|
+
test("2*3");
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|
71
|
+
test("1+2*3");
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|
72
|
+
test("(1+2)*3");
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73
|
+
test("1+3*(2+4)");
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|
74
|
+
}
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|
75
|
+
```
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|
76
|
+
```plain
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|
77
|
+
f=1+2
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78
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+
a=3
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79
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+
f=2*3
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80
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+
a=6
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81
|
+
f=1+2*3
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82
|
+
a=7
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83
|
+
f=(1+2)*3
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84
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+
a=9
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85
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+
f=1+3*(2+4)
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86
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+
a=19
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87
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+
```
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|
88
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+
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1
89
|
参考までに、スタックでベタに逆ポーランド記法利用時はこんな感じかと。
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2
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-
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3
90
|
```c
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4
91
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#include <stdio.h>
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5
92
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#include <stdlib.h>
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1
説明追加
answer
CHANGED
|
@@ -1,4 +1,4 @@
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1
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-
参考までに、逆ポーランド記法利用時はこんな感じかと。
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1
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+
参考までに、スタックでベタに逆ポーランド記法利用時はこんな感じかと。
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2
2
|
|
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3
3
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```c
|
|
4
4
|
#include <stdio.h>
|