C++ 如何从递归函数返回指针，该函数将波兰语表示法转换为反向波兰语表示法？_C++_C_Pointers_Recursion_Postfix Notation

C++ 如何从递归函数返回指针，该函数将波兰语表示法转换为反向波兰语表示法？

c++ c pointers recursion

C++ 如何从递归函数返回指针，该函数将波兰语表示法转换为反向波兰语表示法？,c++,c,pointers,recursion,postfix-notation,C++,C,Pointers,Recursion,Postfix Notation,我的一个朋友被指派编写一个函数，将波兰语表示法转换为反向波兰语表示法，因此在最后一个字符串“sin*-x4+83”应转换为“x4-83+*sin”。有一些约束，例如：在结束时，应该返回一个指针，并在堆上分配内存；函数原型应该只包含一个参数，并且应该是原始字符串。我知道使用静态变量是对人类的可怕犯罪，但尽管如此，我们得出的最好的结论是： char * prefix2postfix( char ppn_in[] ){ if (ppn_in[0] == '\0') { ret

我的一个朋友被指派编写一个函数，将波兰语表示法转换为反向波兰语表示法，因此在最后一个字符串“sin*-x4+83”应转换为“x4-83+*sin”。有一些约束，例如：在结束时，应该返回一个指针，并在堆上分配内存；函数原型应该只包含一个参数，并且应该是原始字符串。我知道使用静态变量是对人类的可怕犯罪，但尽管如此，我们得出的最好的结论是：

char * prefix2postfix( char ppn_in[] ){

    if (ppn_in[0] == '\0') {
        return ppn_in;
    }

    int i = 0; 
    char *left = "",
         *root = "";

    while (ppn_in[i] == ' ') i++;

    if (is_num(ppn_in[i]) || is_var(ppn_in[i])) {
        char tmp[8];
        memset(tmp, '\0', 7);
        tmp[0] = ' ';
        tmp[1] = ppn_in[i];
        root = tmp;
        left = prefix2postfix(ppn_in + i + 1);
    } else if (is_op(ppn_in[i])) {
        switch (ppn_in[i]) {
            case '+':
                root = " +";
                break;
            case '-':
                root = " -";
                break;
            case '*':
                root = " *";
                break;
            case '/':
                root = " /";
                break;
        }
        left = prefix2postfix(ppn_in + i + 1);
    } else if (strncmp(ppn_in, "sqrt", 4) == 0) {
        root = " sqrt";
        left = prefix2postfix(ppn_in + i + 4);
    } else if (strncmp(ppn_in, "sqr", 3) == 0) {
        root = " sqr";
        left = prefix2postfix(ppn_in + i + 3); 
    } else if (strncmp(ppn_in, "sin", 3) == 0) {
        root = " sin";
        left = prefix2postfix(ppn_in + i + 3); 
    } else if (strncmp(ppn_in, "cos", 3) == 0) {
        root = " cos";
        left = prefix2postfix(ppn_in + i + 3);
    } else if (strncmp(ppn_in, "tan", 3) == 0) {
        root = " tan";
        left = prefix2postfix(ppn_in + i + 3);
    } else if (strncmp(ppn_in, "log", 3) == 0) {
        root = " log";
        left = prefix2postfix(ppn_in + i + 3);
    }   


    printf("reached end of the function\n");

    static char buff[200];
    buff[199] = '\0';

    strcpy(buff, left); 
    strcat(buff, root);
    return buff;
}

问题是（这可能不是唯一的一个），我不知道如何正确地结束函数，以使行

printf（“到达函数的结尾\n”）

只被调用一次，在这种情况下，我将能够使用

malloc

，否则我将导致内存泄漏。任何建议都将不胜感激

编辑

我试着自己来做一些事情，我最终会造成内存泄漏吗？（我知道这是非常低效的，但我分配了200字节，将它们复制到一个新的缓冲区，使用该值后，我调用free）

您的解析并不是使用两个分支进行递归，但要回答您的问题，每次都需要malloc返回值，否则调用方将不知道何时释放。我把这个例子作为C代码来保存，因为我在你的问题中没有看到任何C++。

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

int is_num( char v ) { return isdigit( v ); }
int is_var( char v ) { return isalpha( v ); }
int is_op( char v ) { return ispunct( v ); }

char * prefix2postfix( char ppn_in[] ){

    if (ppn_in[0] == '\0') {
        char * ptr = malloc( sizeof( char ) * 1024 );
        ptr[0] = '\0';
        return ptr;
    }

    int i = 0; 
    char tmp[16];
    char *left = 0,
         *root = "";

    while (ppn_in[i] == ' ') i++;

    if ( is_var(ppn_in[i])) {
        tmp[0] = ' ';
        tmp[1] = ppn_in[i];
        char *tptr = &tmp[2];
        while( is_var(ppn_in[i + 1] ) )
            *tptr++ = ppn_in[++i];
        *tptr = '\0';
        root = tmp;
        left = prefix2postfix(ppn_in + i + 1);
    } else if (is_num(ppn_in[i]) || is_var(ppn_in[i])) {
        char tmp[8];
        memset(tmp, '\0', 7);
        tmp[0] = ' ';
        tmp[1] = ppn_in[i];
        root = tmp;
        left = prefix2postfix(ppn_in + i + 1);
    } else if (is_op(ppn_in[i])) {
        switch (ppn_in[i]) {
            case '+':
                root = " +";
                break;
            case '-':
                root = " -";
                break;
            case '*':
                root = " *";
                break;
            case '/':
                root = " /";
                break;
        }
        left = prefix2postfix(ppn_in + i + 1);
    } else if (strncmp(ppn_in, "sqrt", 4) == 0) {
        root = " sqrt";
        left = prefix2postfix(ppn_in + i + 4);
    } else if (strncmp(ppn_in, "sqr", 3) == 0) {
        root = " sqr";
        left = prefix2postfix(ppn_in + i + 3); 
    } else if (strncmp(ppn_in, "sin", 3) == 0) {
        root = " sin";
        left = prefix2postfix(ppn_in + i + 3); 
    } else if (strncmp(ppn_in, "cos", 3) == 0) {
        root = " cos";
        left = prefix2postfix(ppn_in + i + 3);
    } else if (strncmp(ppn_in, "tan", 3) == 0) {
        root = " tan";
        left = prefix2postfix(ppn_in + i + 3);
    } else if (strncmp(ppn_in, "log", 3) == 0) {
        root = " log";
        left = prefix2postfix(ppn_in + i + 3);
    }   


    printf("reached end of the function  '%s'  '%s'\n", left ? left : "", root );

    if( left )
    {
        strcat( left, root );
        return left;
    }
    char * buff = malloc( sizeof( char ) * 1024 );
    strcpy( buff, root );
    return buff;
}

int main( int argc, char ** argv )
{
    if( argc == 2 )
    {
        char * ptr = prefix2postfix( argv[1] );
        printf( "Result: %s\n", ptr );
        free( ptr );
    }
}

您需要传递一个缓冲区来写入结果，或者让它从空闲存储区动态分配。调用者是如何使用它的？@GeorgeHoupis他获取结果并将其分配给一个char指针，使用后他调用“free”是的，你是对的，我从左和右开始，然后删除了右和其他变量。起初我认为这是一个树问题，但我无法解决它，所以我最终得到了我的结果。我说得对吗？在最后一步中为left获取一个缓冲区，并防止到达函数末尾，直到最后一个缓冲区结束？但是什么时候释放left？我在main中释放left，因为它最终由函数返回，或者我重用从递归的较低调用中分配的内容。您在某种程度上是正确的，但是您没有正确解析，因为您正在反转参数，使减法反转。您只想将操作符并置。@orustammanapov我给您提供了另一个您可以研究的解决方案。

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

int is_num( char v ) { return isdigit( v ); }
int is_var( char v ) { return isalpha( v ); }
int is_op( char v ) { return ispunct( v ); }

char * prefix2postfix( char ppn_in[] ){

    if (ppn_in[0] == '\0') {
        char * ptr = malloc( sizeof( char ) * 1024 );
        ptr[0] = '\0';
        return ptr;
    }

    int i = 0; 
    char tmp[16];
    char *left = 0,
         *root = "";

    while (ppn_in[i] == ' ') i++;

    if ( is_var(ppn_in[i])) {
        tmp[0] = ' ';
        tmp[1] = ppn_in[i];
        char *tptr = &tmp[2];
        while( is_var(ppn_in[i + 1] ) )
            *tptr++ = ppn_in[++i];
        *tptr = '\0';
        root = tmp;
        left = prefix2postfix(ppn_in + i + 1);
    } else if (is_num(ppn_in[i]) || is_var(ppn_in[i])) {
        char tmp[8];
        memset(tmp, '\0', 7);
        tmp[0] = ' ';
        tmp[1] = ppn_in[i];
        root = tmp;
        left = prefix2postfix(ppn_in + i + 1);
    } else if (is_op(ppn_in[i])) {
        switch (ppn_in[i]) {
            case '+':
                root = " +";
                break;
            case '-':
                root = " -";
                break;
            case '*':
                root = " *";
                break;
            case '/':
                root = " /";
                break;
        }
        left = prefix2postfix(ppn_in + i + 1);
    } else if (strncmp(ppn_in, "sqrt", 4) == 0) {
        root = " sqrt";
        left = prefix2postfix(ppn_in + i + 4);
    } else if (strncmp(ppn_in, "sqr", 3) == 0) {
        root = " sqr";
        left = prefix2postfix(ppn_in + i + 3); 
    } else if (strncmp(ppn_in, "sin", 3) == 0) {
        root = " sin";
        left = prefix2postfix(ppn_in + i + 3); 
    } else if (strncmp(ppn_in, "cos", 3) == 0) {
        root = " cos";
        left = prefix2postfix(ppn_in + i + 3);
    } else if (strncmp(ppn_in, "tan", 3) == 0) {
        root = " tan";
        left = prefix2postfix(ppn_in + i + 3);
    } else if (strncmp(ppn_in, "log", 3) == 0) {
        root = " log";
        left = prefix2postfix(ppn_in + i + 3);
    }   


    printf("reached end of the function  '%s'  '%s'\n", left ? left : "", root );

    if( left )
    {
        strcat( left, root );
        return left;
    }
    char * buff = malloc( sizeof( char ) * 1024 );
    strcpy( buff, root );
    return buff;
}

int main( int argc, char ** argv )
{
    if( argc == 2 )
    {
        char * ptr = prefix2postfix( argv[1] );
        printf( "Result: %s\n", ptr );
        free( ptr );
    }
}

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

int is_binary_operator( const char * token )
{
    return ( ( *token == '+' ) ||
             ( *token == '*' ) ||
             ( *token == '-' ) ||
             ( *token == '/' ) );
}

int is_unary_operator( const char * token )
{
    return ( !strcmp( token, "sin" ) ||
             !strcmp( token, "cos" ) ||
             !strcmp( token, "tan" ) ||
             !strcmp( token, "exp" ) ||
             !strcmp( token, "log" ) ||
             !strcmp( token, "ln" ) ||
             !strcmp( token, "sqrt" ) ||
             !strcmp( token, "sqr" ) );
}

int reverse_polish_arranger( char ** tokens, size_t cnt, size_t iter/*, size_t depth*/ )
{
    /*printf( "%*sPeeking at %s\n", depth << 1, "", tokens[iter] );*/
    if( is_binary_operator( tokens[iter] ) )
    {
        int next_pos = reverse_polish_arranger( tokens, cnt, iter + 1 /*, depth + 1 */);
        int end_pos = reverse_polish_arranger( tokens, cnt, next_pos + 1 /*, depth + 1 */);
        char * ptr = tokens[iter];
        memmove( tokens + iter, tokens + iter + 1, sizeof( char * ) * ( end_pos - iter  ) );
        tokens[end_pos] = ptr;
        return end_pos;
    }
    else if( is_unary_operator( tokens[iter] ) )
    {
        int end_pos = reverse_polish_arranger( tokens, cnt, iter + 1 /*, depth + 1*/ );
        char * ptr = tokens[iter];
        memmove( tokens + iter, tokens + iter + 1, sizeof( char * ) * ( end_pos - iter ) );
        tokens[end_pos] = ptr;
        return end_pos;
    }
    else
        return iter;
}

char * reverse_polish( const char * expr )
{
    size_t index;
    size_t push_index = 0;
    size_t len = 0;
    char * result = 0;
    char * buffer = 0;
    char ** pointers = 0;

    /* Prepare the workspace. */
    if( !expr )
        return 0;
    len = strlen( expr );
    if( !( result = malloc( sizeof( char ) * len ) ) )
        return 0;
    if( !( buffer = malloc( sizeof( char ) * len ) ) )
    {
        free( result );
        return 0;
    }
    if( !( pointers = malloc( sizeof( char * ) * len ) ) )
    {
        free( result );
        free( buffer );
        return 0;
    }
    strcpy( buffer, expr );
    memset( pointers, 0, sizeof( char * ) * len );

    /* Cheap tokenize using space a delimiter. */
    pointers[push_index++] = buffer;
    for( index = 0; index < len; ++index )
        if( buffer[index] == ' ' )
            buffer[index++] = '\0', pointers[push_index++] = &buffer[index];

    /* printf( "Before:\n" );
     * for( index = 0; index < push_index; ++index )
     *    printf( "Token %3d: %s\n", index, pointers[index] );
     */

    /* Do the conversion. */
    reverse_polish_arranger( pointers, push_index, 0/*, 0*/ );

    /*
     * printf( "After:\n" );
     * for( index = 0; index < push_index; ++index )
     *    printf( "Token %3d: %s\n", index, pointers[index] );
     */

    /* Prepare output buffer. */
    result[0] = '\0';
    for( index = 0; index < push_index; ++index )
        strcat( strcat( result, pointers[index] ), " " );

    /* Cleanup temporary workspace */
    free( pointers );
    free( buffer );

    /* Return new string. */
    return result;
}

int main( int argc, char ** argv )
{
    if( argc == 2 )
    {
        char * reversed = reverse_polish( argv[1] );
        if( reversed )
        {
            printf( "Reverse: %s\n", reversed );
            free( reversed );
        }
        else
            printf( "Failed to reverse: %s\n", argv[1] );
    }
    return 0;
}

./a.out "sin * - x 4 + 8 3"
Reverse: x 4 - 8 3 + * sin