char 之前的预期表达式

Question

我编写了这个程序，它询问用户一个名字，然后打印这个名字。详细步骤如下：

• 询问用户名称（即句子）的字符数。它包含空格和终止符\0，然后存储它；
• 它使用num_char地址创建一个内存块，并将第一项的地址存储在ptr中；
• 在else部分中，声明了一个未知大小的数组，用于存储名称，并将其首地址分配给ptr；
• 然后为数组分配大小num_char；

代码如下：

#include <stdio.h>
#include <stdlib.h>
//an attempt at a program that asks for a name, stores it in an array and then prints it

int main() {
    int num_char;
    char *ptr;

    printf("input number of characters of your name (spaces and terminator character included): ");

    scanf("%d", &num_char);

    ptr = char *malloc(num_char * sizeof(char)); //creates a block of mem to store the input name, with the same size, and then returns the adress of the beginning of the block to ptr

    if (ptr == NULL) {
        printf("allocation not possible");
    } else {
        ptr = char name[]; //ptr stores the adress of the first char in string

        char name[num_char], //declaration of an array with num_char elements

        printf("input name: ");

        scanf("%s", name);
        printf("input name was: %s", name);
    }
    return 0;
} 

但是我得到三个编译错误：

• “'char' 之前的预期表达式”ptr = char *malloc(num_char * sizeof(char) ); 和 ptr = char name[];
• “预期的声明说明符或字符串常量之前的'...'”printf("input name: ");

我是一名刚开始学习 C 和一般编程的大学生，因此非常感谢详细解释任何类型的错误以及如何修复它：)

Answer 1

这些是语法错误，因为：

• ptr = char *malloc(num_char * sizeof(char)); 应该是 ptr = (char *)malloc(num_char * sizeof(char)); 或只是 ptr = malloc(num_char * sizeof(char));。

• ptr = char name[]; 毫无意义。您根本不需要更改ptr。

• char name[num_char], 是一个 C99 可变长度数组定义，但有一个尾随 ,，因此下一行应该有另一个定义，因此出现第二个错误。

您必须选择使用malloc() 从堆中分配内存或将数组定义为局部变量。

这是修改后的版本：

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

int main() {
    int num_char;
    char *ptr;

    printf("input number of characters of your name (spaces and terminator character included): ");

    if (scanf("%d", &num_char) != 1 || num_char <= 1) {
        printf("invalid input\n");
        return 1;
    }
    /* discard the rest of the input including the newline */
    int c;
    while ((c = getchar()) != EOF && c != '\n')
        continue;

    //allocate a block of memory to store the name
    ptr = malloc(num_char * sizeof(char));

    if (ptr == NULL) {
        printf("allocation not possible");
    } else {
        printf("input name: ");

        if (scanf("%[^\n]", ptr) == 1)
            printf("input name was: %s\n", name);
        else
            printf("no input was given\n");

        free(ptr);
    }
    return 0;
} 

还要注意，如果输入有更多的非白色字节，scanf() 可能会读取超过num_char 字节，因此给scanf() 提供有关数组大小的信息会更安全，但并非微不足道：

    if (ptr != NULL) {
        char format[32];
        snprintf(format, sizeof format, "%%%d[^\n]", num_char - 1);
        printf("input name: ");
        if (scanf(format, ptr) == 1) {
            printf("input name was: %s\n", name);
            // should discard the rest of the line and the newline
        } else {
            printf("no input was given\n");
        }
    }

Answer 2

正如我在 cmets 中所说，这是一组奇怪的要求，我假设是一个作业，这里或多或少是你应该使用 cmets 做的：

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

void clear_buffer()
{ // routine to clear standard input
    int c;
    while ((c = getchar()) != '\n' && c != EOF){}
    if (c == EOF)
    {
        exit(EXIT_FAILURE);
    }
}

int main()
{
    int num_char;
    char *ptr;

    printf("input number of characters of your name (spaces and terminator character included): ");
    // less than 9 + 1 null byte characters for the name not allowed
    while (scanf("%d", &num_char) != 1 || num_char < 10)
    { 
        printf("Bad input, try again: ");
        clear_buffer();
    }
    ptr = malloc(num_char); // removed char*, you may wanted a cast, 
                            // but even that is unneeded,
                            // The size of char is always 1 byte
                            // I still don't understand why this is asked, but well...
    if (ptr == NULL)
    {
        perror("allocation not possible"); // perror prints the apropriate error
    }
    else
    {
        char name[num_char]; // declaration of an array with num_char elements
        free(ptr);           // to make ptr point to another memory location, it must be freed
        ptr = name;          // ptr now points to the first element of the name array

        clear_buffer();
        printf("What's the name ? ");
        // read name with spaces from standard input
        if (fgets(name, num_char, stdin))
        {                   
            name[strcspn(name, "\n")] = '\0'; // remove \n 
            printf("input name was: %s", ptr); // print name through ptr 
        }  
        else
        {
            puts("Bad input");
        }                    
       
    }
    return EXIT_SUCCESS;
}

等待模棱两可的解释...