Question

我有一个结构定义，需要由多个不同的函数使用。在我的struct定义中，我有一个大小为“len”的数组，这个len变量是argv [1]中字符串的长度。正如您可能看到的，我需要这个变量，但我不能将结构定义放在main之外，否则我会丢失该变量。但我确实需要将它放在我的一个函数的原型之前。这个问题的解决方案是什么？这是我的代码：

void randomize( type_darwin *darwin, type_monkey *monkey );

int main( int argc, char *argv[] )
{
    if ( argc < 2 )
    {
        printf("Error! Arguments are of form: ./%s [string to parse]\nBe sure your string is surrounded by double quotes.\n", argv[0]);
        return 0;
    }

    int len = strlen(argv[1]);              // length of string
    const char *string = argv[1];           // make string a constant

    // define two structs, one for the Darwinian algorithm and one for the monkey bashing algorithm
    // lock determines whether that element in sentence is locked (in programming terms, if it should now be a constant)
    typedef struct darwin
    {
        char sentence[len];
        int lock[len];  // 0 defines not locked. Nonzero defines locked.
    } type_darwin;

    typedef struct monkey
    {
        char sentence[len];
        int lock;   // 0 defines entire array not locked. Nonzero defines locked.
    } type_monkey;

Answer 1

结构需要具有一致的编译时定义才能以这种方式使用。你最好使用指针而不是静态数组并动态地为数组分配空间。然后，您需要使结构的长度部分。

typedef struct darwin
{
    int len;
    char *sentence;
    int *lock;  // 0 defines not locked. Nonzero defines locked.
} type_darwin;

typedef struct monkey
{
    int len;
    char *sentence;
    int lock;   // 0 defines entire array not locked. Nonzero defines locked.
} type_monkey;


int main(int argc, char *argv[] )
{
    if ( argc < 2 )
    {
        printf("Error! Arguments are of form: ./%s [string to parse]\nBe sure your string is surrounded by double quotes.\n", argv[0]);
        return 0;
    }

    int len = strlen(argv[1]);              // length of string
    const char *string = argv[1];           // make string a constant

    type_darwin my_darwin;
    my_darwin.len = len;
    my_darwin.sentence = malloc(len + 1);
    my_darwin.lock = malloc((len + 1) * sizeof(int));

    type_monkey my_monkey;
    my_monkey.len = len;
    my_monkey.sentence = malloc(len + 1);

    ...

}

在函数原型之前放置struct定义，但struct需要来自argv的信息

1 个答案: