我的函数不起作用，我也不明白为什么..in C

Question

我正在处理一个问题，保存商店库存的证据，这里有一个部分，它被要求得到产品的代码，我必须通过倒换产品的名称来得到。

例如：

name of the product : snack
codebar: kcans

而且每件事情都很好，直到它到达了与strcpy的线。

菜单不是英文的，但是第一个菜单得到输入，第二个菜单显示输出。

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


typedef struct Product
{
    char productName[30];
    char productCode[30];
    int price;
    int stock;
    int date[3];
}P;

char* reverse(char *name)
{
    char aux[30];
    int i,j;

    for(i = strlen(name) - 1, j = 0; i >= 0; i--,j++)
    {
        aux[j] = name[i];
    }
    aux[j] = '\0';

    return aux[30];
}


int nameValidation(char *name)
{
    int i;
    for(i = 0; i < strlen(name); i++)
    {
        if(name[i] >= 'a' && name[i] <= 'z');
        else
        {
            return 0;
            break;
        }
    }
    if(i == strlen(name))
    return 1;
}

void input(P *product, int *n)
{
    (*n)++;
    do
    {
        printf("Product name: ");
        scanf("%s", (product + *n)->productName);

    } while (!nameValidation((product + *n)->productName));


    printf("Price: ");
    scanf("%d", &(product + *n)->price);

    printf("Quantity: ");
    scanf("%d", &(product + *n)->sotck);


    do
    {
        printf("Expiration day: ");
        scanf("%d", &(product + *n)->date[0]);

        if((product + *n)->date[0] >= 1 && (product + *n)->date[0] <= 30)
            break;
    } while (1);
   
    
    do
    {
        printf("Expiration month: ");
        scanf("%d", &(product + *n)->date[1]);

        if((product + *n)->date[1] >= 1 && (product + *n)->date[1] <= 12)
            break;
    } while (1);
    

    (product + *n)->date[2] = 2024;
    printf("Expiration Year: %d\n", (product + *n)->date[2]);

    strcpy((product + *n)->productCode, reverse((product + *n)->productName));
    printf("Product Code: %5s\n", (product + *n)->productCode);

}


void output(P *product, int n)
{
    int i;
    for(i = 0; i <= n; i++)
    {
        printf("Product name: %10s ", (product + i)->productName);

        printf("Price: %5d ", (product + i)->price);

        printf("Stock:%5d\n", (product + i)->stock);
    }

}


int main()
{
    int option;
    int n;
    P product[100];

    n = -1;
    do
    {
        printf("\n0.Iesire\n");
        printf("1.Citirea unui produs de la tastatura\n");
        printf("2.Afisarea produselor citite\n");
        printf("3.Afisarea produselor cu un pret mai mic dat de utilizator\n");
        printf("4.Sa se afiseze toate produsele care au lungimea numelui egala cu un numar par\n");
        printf("5.Afisarea primului produs care este egal cu un nume dat de utilizator\n");
        printf("6.Sa se realizeze produsul cifrelor fiecarui numar de bucati al produselor, apoi sa se copieze fiecare produs intr-un vector si sa se afiseze vectorul.\n");
        printf("7.Copierea tuturor codurilor produselor intr-un vector de sir de caractere, apoi afisare vectorului\n");
        printf("8.Copierea pretului produselor intr-un vector apoi afisarea vectorului in ordine crescatoare\n");
        printf("9.Sa se transforme prima litera din codul produsului in litera mare daca aceasta este vocala.Nu se vor folosii functii predefinite in schimbarea literei\n");
        printf("10.Sa se afiseze forma binara a numarului de bucati ale produselor care au numarul de bucati mai mic decat 100 .\n");
        printf("11.Sa se afiseze toate produsele care urmeaza sa expire in urmatoarele 10 zile.Utilizator va preciza de la tastatura ziua si luna curenta.\n");
        printf("12.Sa se pun pe ultimul rand al unei matrici de caractere de ordin dat de numarul de produse, primul caracter din fieacre nume de produs,pe restul '.'\n");
        printf("13.Stabilirea unui nou pret pt toate produsele daca utilizatorul doreste.Astfel utilizator va stabili procentajul cu care se vor scumpi produsele.\n");
        printf("14.Sa se stearga un produs precizat de utilizator prin numele acestuia\n");
        printf("Introduceti o optiune: ");
        scanf("%d", &optiune);

        switch (option)
        {
        case 1:
            input(product, &n);
            break;
        case 2:
            output(product, n);
            break;
        case 3:
            
            break;
        case 4:
            /* code */
            break;
        case 5:
            /* code */
            break;
        case 6:
            /* code */
            break;
        case 7:
            /* code */
            break;
        case 8:
            /* code */
            break;
        case 9:
            /* code */
            break;
        case 10:
            /* code */
            break;
        case 11:
            /* code */
            break;
        case 12:
            /* code */
            break;
        case 13:
            /* code */
            break;
        case 14:
            /* code */
            break;
        case 0:
            exit(1);
            break;
        
        default:
            printf("Introduceti o optiune valida");
            break;
        }

    } while (option != 0);
    
    return 0;
}

Answer 1

你不应该浪费时间输入15个‘提示’和15个大小写标签，就好像添加更多的代码会让事情顺利进行一样。我强烈建议删除该代码(或注释掉这些行)，并将注意力集中在需要构建的小实用程序上。

这里有一个“固定”版本的reverse()，让您开始工作。请增量地工作，测试添加的每一行额外代码，以及您所做的每一个微小的更改。山不会自上而下生长..。

char *reverse( char *str ) {
    for( int i = 0, j = strlen( str ) - 1; i <= j; i++, j-- ) {
        char tmp = str[i];
        str[i] = str[j];
        str[j] = tmp;
    }
    return str;
}

int main() {
    char *str[] = { "the fish", "goats are good" };

    printf( "%s", str[0] );
    printf( " == %s\n", reverse( str[0] ) );

    printf( "%s", str[1] );
    printf( " == %s\n", reverse( str[1] ) );

    return 0;
}

输出：

the fish == hsif eht
goats are good == doog era staog

编辑：

由于对这个答案的批评，我觉得有责任补充一下这一点。

由于此函数反转字符串“就位”，因此需要修改以下代码行：

strcpy( (product + *n)->productCode, reverse((product + *n)->productName) );

至

reverse( strcpy( (product + *n)->productCode, (product + *n)->productName) );

Answer 2

我正在发布这个公认的原始代码的“代码转储”修订版(附带一些注释)，以演示“大图”草图如何与增量开发共存。请从main()的入口点到exit() of case 0:来学习这个例子。注意更短的名称是如何“澄清”函数中的代码的。当您想要“复制/粘贴/调整”某些代码行时，请问问自己，是否可以通过分离一个单独的函数来更好地服务这些功能。

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

typedef struct Product {
    char name[30]; // meaningful but shorter name
    char code[30]; // meaningful but shorter name
    int price;
    int stock;
    int exp_yr, exp_mo, exp_dy; // no confusion
} P_t; // conventional to use "_t" with tokens 'typedef'd

// primative 'helper' functions
char *getString( char *prompt, char *to, int len ) {
    printf( "%s", prompt );
    fgets( to, len, stdin );
    to[ strlen(to) - 1 ] = '\0'; // get rid of '\n'
    return to;
}

int getInt( char *prompt ) {
    char buf[32];
    return atoi( getString( prompt, buf, sizeof buf ) );
}

char *reverse( char *str ) { // "in place" string reversal
    int lft = 0;
    int rgt = strlen( str ) - 1;

    for( ; lft <= rgt; lft++, rgt-- ) {
        char tmp = str[ lft ];
        str[ lft ] = str[ rgt ];
        str[ rgt ] = tmp;
    }
    return str;
}

int nameValid( char *str ) { // check for bad chars
    if( *str == '\0' ) return 0; // At least one char required

    char *permitted = "- abcdefghijklmnopqrstuvwxyz"; // extra acceptable chars

    return strspn( str, permitted ) == strlen( str ); // true if all chars okay
}

void input( P_t *pp ) { // fill-in ONE record
    do {
        getString( "Product name: ", pp->name, sizeof pp->name );
    } while( !nameValid( pp->name ) );

    reverse( strcpy( pp->code, pp->name ) );
    printf( "Product Code: %s\n", pp->code );

    pp->price = getInt( "Price: " );
    pp->stock = getInt( "In stock: " );

    // Not good, but sufficient
    while( ( pp->exp_dy = getInt( "Exp. day: " ) ) < 1 || pp->exp_dy > 31 ) {} // loop

    while( ( pp->exp_mo = getInt( "Exp. month: " ) ) < 1 || pp->exp_mo > 12 ) {} // loop

    pp->exp_yr = getInt( "Exp. year: " );
}

// Renamed to show outputs multiple records
// Equivalent, but easier to think of as array
void outputAll( P_t prod[], int n ) {

    for( int i = 0; i < n; i++ ) { // NB: changed '<=' to '<'
        P_t *pp = &prod[ i ]; // less verbose

        printf( "Product name: %10s ", pp->name );
        printf( "Price: %5d ", pp->price );
        printf( "Stock:%5d", pp->stock );
        printf( "\n" ); // Clearly 'end of line'
    }
}

int main() {
    P_t product[100]; // uninitialised
    int count = 0;

    memset( product, 0, sizeof product ); // Now initialised

    for( ;; ) {
        printf(
            "\n0.Iesire\n" );
        printf(
            "1.Citirea unui produs de la tastatura\n" );
        printf(
            "2.Afisarea produselor citite\n" );
/*      printf(
            "3.Afisarea produselor cu un pret mai mic dat de utilizator\n" );
        printf(
            "4.Sa se afiseze toate produsele care au lungimea numelui egala cu un numar par\n" );
        printf(
            "5.Afisarea primului produs care este egal cu un nume dat de utilizator\n" );
        printf(
            "6.Sa se realizeze produsul cifrelor fiecarui numar de bucati al produselor, "
            "apoi sa se copieze fiecare produs intr-un vector si sa se afiseze vectorul.\n" );
        printf(
            "7.Copierea tuturor codurilor produselor intr-un vector de sir de caractere, "
            "apoi afisare vectorului\n" );
        printf(
            "8.Copierea pretului produselor intr-un vector "
            "apoi afisarea vectorului in ordine crescatoare\n" );
        printf(
            "9.Sa se transforme prima litera din codul "
            "produsului in litera mare daca aceasta este vocala. "
            "Nu se vor folosii functii predefinite in schimbarea literei\n" );
        printf(
            "10.Sa se afiseze forma binara a numarului de bucati "
            "ale produselor care au numarul de bucati mai mic decat 100 .\n" );
        printf(
            "11.Sa se afiseze toate produsele care urmeaza sa expire in urmatoarele 10 zile. "
            "Utilizator va preciza de la tastatura ziua si luna curenta.\n" );
        printf(
            "12.Sa se pun pe ultimul rand al unei matrici de caractere de ordin "
            "dat de numarul de produse, primul caracter "
            "din fieacre nume de produs,pe restul '.'\n" );
        printf(
            "13.Stabilirea unui nou pret pt toate produsele daca utilizatorul "
            "doreste.Astfel utilizator va stabili procentajul cu "
             "care se vor scumpi produsele.\n" );
        printf(
            "14.Sa se stearga un produs precizat de utilizator prin numele acestuia\n" );
*/

        switch ( getInt( "Introduceti o optiune: " ) ) {
            case 0:
                exit( 1 ); // Would 'return 0' be better?
                break;

            case 1:
                // increment it here, instead of using pointer there.
                // pass only one element to be filled in.
                input( &product[ count++ ] );
                break;

            case 2:
                outputAll( product, count );
                break;

// ToDo:
//          case 3: break;
//          case 4: break;
//          case 5: break;
//          case 6: break;
//          case 7: break;
//          case 8: break;
//          case 9: break;
//          case 10: break;
//          case 11: break;
//          case 12: break;
//          case 13: break;
//          case 14: break;
// End ToDo

            default:
                printf("Introduceti o optiune valida");
                break;
        }
    }
    
    return 0;
}