字符串替换件 [英] replacing pieces of string
问题描述
我在做类似Excel中，我有这样的事情：
1 2 3
A1 B1 C1
它取代了指定的内容，其中A1取代1的内容的内容。
B1取代了2内容...等...
我使用的是多维数组，我真的很喜欢这样的事情：
INT偏移= 0，readCharCount;
而（sscanf的（MATRIS [I] [C] +偏移，％C％D％* C％N，＆安培;山坳，和放大器; linha，＆安培; readCharCount）== 2）{
//的printf（％C，％d个\\ N，山坳，linha）;
//的strcpy（MATRIS [I] [C]，MATRIS [linha-1] [山口 - 'A']）; 胶印+ = readCharCount;
//输出（{％C，％D}，山坳，linha）;
//的printf（\\ n）;
}
但是，当我有A1 + B1 + C1和另外的事情，我不能代替的总含量，因为其他的引用将被删除......
所以，在细胞内，A1 + B1 + C1，我想改变为B1指定的内容....我想有这样的：
此 - ＆GT; A1 + B1 + C1至 - ＆GT; 1 + 2 + 3
...
感谢。
您可能只是重复使用该<一个href=\"http://stackoverflow.com/questions/17112494/how-to-expand-environment-variables-in-ini-files-using-boost/17128601#17128601\">c++解决方案（通过硬编码的char * 不是替换通用迭代器）。
我给它一个旋转。不过，我想给一个警告：它看起来像你想实现一个前pression解析器。我强烈建议你为
- handroll一个（递归下降）分析器
- 使用柔性/野牛（或法/ yacc的）
所以你不要画在自己容易出错的文本处理C的一个尴尬的角落。
修改我重写 C程序使用C ++;你可以看到它的此处工作现场。
修改2：纯C C程序的另一种修正： 的http：/ /ideone.com/ExnufJ 更新，现在支持迭代扩展，太的
答案只是关注自身与纯C的方法：
那么，让我们开始吧。我以为一个样本的S preadsheet（它可能包含数字而不是字符串）：
为const char *细胞[] [4] = {
/* A B C D */
{的，懒惰，牛，跳楼} / * 1 * /
{超过，中，快速，棕色} / * 2 * /
{纸，套餐，捆绑，向上} / * 3 * /
{与，银，白，冬天} / * 4 * /
{说，融，狐狸精，弹簧} / * 5 * /
};使用只有两个助手：
为const char * get_cell_value（为const char * coordinate_b，为const char * coordinate_e）;
字符* expand_cell_references（为const char *楼为const char * const的L，字符* O）; / *魔引擎* /我们可以写出下面的演示程序：
INT的main（）
{
为const char []中=的C2 D2 C5 D1 A2 B2 B1的狗！ 烧焦出[1024] = {0};
expand_cell_references（IN，IN +的strlen（中），淘汰）;
看跌期权（出）; / *敏捷的棕色狐狸跳过了懒狗！ * / 返回0;
}打印出公知的测试短语按照注释。现在，
get_cell_value是非常简单的：为const char * get_cell_value（为const char * coordinate_b，为const char * coordinate_e）
{
为size_t山坳= 0，行= 0;
为const char *它;
对于（IT = coordinate_b;！它= coordinate_e ++吧）
{
如果（*它＆GT; ='A'和;＆放大器; *它＆LT; ='Z'）
COL = 26 * COL +（*吧 - 'A'）;
如果（*它＆GT; ='0'和;＆放大器; *它＆LT; =9）
行= 10 *行+（*吧 - '0'）; / *或使用的atoi和朋友* /
}
行 - ; / * 1，基于行NUMS在Excel * / 回报细胞[行] [COL]; / *在Excel中1为基础的索引* /
}和
expand_cell_references是稍微有点复杂，是一个简单的DFA解析器：的char * expand_cell_references（为const char *楼为const char * const的L，字符* O）
{
枚举parser_state {
其他，
in_coord_col，
in_coord_row
}状态=等; 为坐标/ *临时存储被解析：* /
焦炭ACCUM [16] = {0};
字符* accit = ACCUM;
而（F！= 1）
{
开关（州）/ *假人，过渡的fallthrough订单流，现在* /
{
其他情况：
*（accit = ACCUM）= 0; / *复位累加器* /
而（F = L＆安培;！＆安培;！（* F＆GT; ='A'和;＆放大器; * F＆LT; ='Z'））
*○++ = *˚F++;
/ * * fallthrough /
案例in_coord_col：
而（F = L＆安培;！＆放大器; * F＆GT; ='A'和;＆放大器; * F＆LT; ='Z'）
* accit ++ = *˚F++;
/ * * fallthrough /
案例in_coord_row：
{
为const char *扩大= ACCUM;
如果（F = L＆放大器;！＆放大器; * f是氟烃基; ='0'和;＆放大器; * F＆下; ='9'）
{
而中（f = L＆放大器;！＆放大器; * f是氟烃基; ='0'和;＆放大器; * F＆下; ='9'）
* accit ++ = *˚F++;
扩展= get_cell_value（ACCUM，accit）;
}
其他
{
* accit = 0;
}
而（*展开）
*○++ = *扩大++;
继续; / *状态=等; * /
}
}
}
返回O;
}我花了一些快捷键那里，因为这个语法是如此的简约，但它应该给你从哪里开始正确的想法。
请参阅现场演示这里 http://ideone.com/kS7XqB 等等你可以用它自己玩。请注意，我说的调试（断言）在
get_cell_value功能，所以你不小心引用越界的索引。i'm doing something like excel, i have something like this:
1 2 3 A1 B1 C1where it replaces the content for specified content, where A1 replaces the content for 1. B1 replaces the content of 2...and etc...
i'm using a multidimensional array, and i do the things like this:
int offset = 0, readCharCount; while(sscanf(matris[i][c] + offset, "%c%d%*c%n", &col, &linha, &readCharCount) == 2){ //printf("%c, %d\n", col, linha); //strcpy(matris[i][c], matris[linha-1][col - 'A']); offset += readCharCount; //printf(" {%c, %d}", col, linha); //printf("\n"); }But when i have A1+B1+C1 and another things, i cant replace the total content, because other references will be removed....
So, at the cell, A1+B1+C1, i wanna change B1 for the content specified....i wanna have like this:
This -> A1+B1+C1 to -> 1+2+3....
Thanks.
解决方案You might just reuse this c++ solution (replacing the generic iterators by hardcoding
char*instead).I gave it a whirl. However, I wish to give a warning: it looks like you're trying to implement an expression parser. I'd strongly advise you to either
- handroll a (recursive descent) parser
- use flex/bison (or lex/yacc)
so you don't paint yourself in an awkward corner of error-prone text-handling in C.
Edit: I rewrote your C program using C++; you can see it working live here.
Edit 2: Another fixup of your C program in pure C: http://ideone.com/ExnufJ updated to support iterative expansions now, too
The answer just concerns itself with the pure C approach:
So, let's get started. I assumed a sample "spreadsheet" (it could contain numbers instead of strings):
const char* cells[][4] = { /* A B C D */ { "the" , "lazy" , "cow" , "jumped" }, /* 1 */ { "over" , "the" , "quick", "brown" }, /* 2 */ { "paper", "packages", "tied" , "up" }, /* 3 */ { "with" , "silver" , "white", "winters" }, /* 4 */ { "that" , "melt" , "fox" , "springs" }, /* 5 */ };Using just two helpers:
const char* get_cell_value(const char* coordinate_b, const char* coordinate_e); char* expand_cell_references(const char* f, const char* const l, char* o); /*the magic engine*/we can write the following demo program:
int main() { const char in[] = "The C2 D2 C5 D1 A2 B2 B1 dog!"; char out[1024] = {0}; expand_cell_references(in, in+strlen(in), out); puts(out); /* "The quick brown fox jumped over the lazy dog!" */ return 0; }which prints the well-known test phrase as per the comment. Now,
get_cell_valueis really simple:const char* get_cell_value(const char* coordinate_b, const char* coordinate_e) { size_t col = 0, row = 0; const char* it; for (it=coordinate_b; it != coordinate_e; ++it) { if (*it >= 'A' && *it <= 'Z') col = 26*col + (*it - 'A'); if (*it >= '0' && *it <= '9') row = 10*row + (*it - '0'); /* or use atoi and friends */ } row--; /* 1-based row nums in Excel */ return cells[row][col]; /* 1-based indexes in Excel */ }And
expand_cell_referencesis slightly more involved, being a simple DFA parser:char* expand_cell_references(const char* f, const char* const l, char* o) { enum parser_state { other, in_coord_col, in_coord_row } state = other; /*temporary storage for coordinates being parsed:*/ char accum[16] = {0}; char* accit = accum; while (f!=l) { switch(state) /*dummy, the transitions flow in fallthrough order for now*/ { case other: *(accit = accum) = 0; /*reset the accumulator*/ while (f!=l && !(*f>='A' && *f<='Z')) *o++ = *f++; /*fallthrough*/ case in_coord_col: while (f!=l && *f>='A' && *f<='Z') *accit++ = *f++; /*fallthrough*/ case in_coord_row: { const char* expanded = accum; if (f!=l && *f>='0' && *f<='9') { while (f!=l && *f>='0' && *f<='9') *accit++ = *f++; expanded = get_cell_value(accum, accit); } else { *accit = 0; } while (*expanded) *o++ = *expanded++; continue; /*state = other;*/ } } } return o; }I took some shortcuts there, because this grammar is so minimalist, but it should give you a proper idea of where to start.
See a live demo here http://ideone.com/kS7XqB so you can play with it yourself. Note that I added debugging (asserts) to the
get_cell_valuefunction so you don't accidentally reference out-of-bounds indexes.
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