在函式中使用字串陣列

我正在練習字串陣列，但似乎我不明白這是如何作業的，有人可以向我解釋制定此代碼的正確方法是什么嗎？

代碼的目的：將 2 個字串分配給字串陣列并使用函式列印它們。

錯誤： 編譯器沒有發現任何錯誤，但我在終端中根本沒有得到任何輸出。

#include <stdio.h>

#define MAX 100

void function(char **);

int main()
{    
    char *a[MAX]; /*array of max 100 strings*/
    a[0] = "test0";
    function(&a[MAX]);    
    return 0;
}

void function(char *a[MAX])
{    
    a[1] = "test1";
    printf("%s",*a[1]);
    printf("%s",*a[0]);
}

uj5u.com熱心網友回復：

編譯器沒有發現任何錯誤，但我根本沒有在終端中得到任何輸出。

那是因為您傳遞的地址不是您的函式所期望的。

function(&a[MAX]);

您MAX在 array中的 index 處傳遞專案的地址a，但您的函式將其解釋為整個陣列的地址。你應該寫：

function(a)

似乎您已經將引數的宣告方式與其在函式呼叫中的使用方式混淆了。當你宣告一個函式時，你必須指定引數的型別，所以你說例如char *s[]表示一個指向字符的指標陣列。但是當你呼叫一個函式時，編譯器已經知道該函式期望的是什么型別，所以你只需要傳遞一個與宣告型別匹配的變數即可。此背景關系中的方括號，如a[MAX]，被解釋為選擇陣列中的一個元素的下標。

旁白：就風格而言，function是一個糟糕的函式名稱。我知道這只是一個很小的測驗程式，所以沒什么大不了的，但盡量養成給事物命名的習慣。

還：

printf("%s",*a[1]);
printf("%s",*a[0]);

在這里，您正在訪問索引 1 處的專案，但您尚未在那里存盤任何內容。這不是一個好計劃。洗掉第一行，或更改代碼以在索引 1 中存盤某些內容。

此外，您不需要取消參考元素。陣列中的每個元素都是一個字符陣列，并且需要printf()一個 type 值char *，這就是每個元素的值a[n]。所以只需使用：

printf("%s", a[1]);
printf("%s", a[0]);

uj5u.com熱心網友回復：

您似乎對正在使用的指標及其正確型別有些困惑。希望下面的程式可以清除一些事情。解釋在評論中：

#include <stdio.h>

#define MAX 100

void function(char* (*aPtr)[MAX])
{
    // Since aPtr is a pointer, we need to derefecence it with * before assigning
    // string literals
    (*aPtr)[1] = "test1";
    printf("In function\n");
    printf("%s\n",(*aPtr)[0]);
    printf("%s\n\n",(*aPtr)[1]);
}

void function2(char** a)
{
    // Remember, * and [] can both be used to dereference in C
    // This is equivalent to *(a 2) = "test2";
    a[2] = "test2";
    printf("In function2\n");
    printf("%s\n",a[0]);
    printf("%s\n",a[1]);
    printf("%s\n\n",a[2]);
}

int main(void)
{
    // This is an array of MAX char pointers. Just on declaration, each pointer in the array
    // points to nothing. You can point them to a string literal (as you have done), or use
    // something like malloc to allocate space for each one
    char *a[MAX];
    // This is a pointer to an array of MAX char pointers. This is the type you're trying to
    // pass to `function`, which is not used correctly
    char* (*aPtr)[MAX] = &a;
    // I've included these prints so you can see the difference between `a` and `aPtr` via pointer
    // arithmetic. In this printf, `a` "decays" to a pointer to the first element in the array.
    // The first element of the array is a `char*`, so a pointer to that is a `char**`, so
    // that's what `a` is in this context, pointing to a[0]. What the actual address is here isn't
    // important, just think of this as offset 0
    printf("pointer to first element      = %p\n", (void*)a);
    // This is the explicit address of the first element in the array. Notice how it is
    // identical to the address of the array above.
    printf("address of first element      = %p\n", ((void*)&(a[0])));
    // This illustrates the "decay" concept again. `a` decays to a `char**`, and  1 on that
    // shows pointer arithmetic. On this architecture, pointers are 8 bytes, so a 1
    // advances the pointer by 8 bytes, which points to the 2nd element in the array.
    // This output will be the offset  8
    printf("pointer to second element     = %p\n", (void*)(a 1));
    // This shows the address of aPtr. Remember, this is a pointer to an array of char
    // pointers 100 large. This base address is also the address of where the array starts,
    // so it will be identical to offset
    printf("address of array              = %p\n", (void*)aPtr);
    // This is where the different pointer types are illustrated. Since char pointers are 8
    // bytes on this architecture, an array of 100 of them is 8*100 = 800 bytes. So aPtr 1
    // here performs pointer arithmetic on that type, meaning it advances the pointer 800
    // bytes. The print out here will be offset  800.
    printf("pointer to next array element = %p\n\n", (void*)(aPtr 1));

    // assign a[0] to point to the string literal "test0"
    a[0] = "test0";

    // This is what you're trying to do (but doing incorrectly) with your function call.
    // This passes the address of `a`, a pointer to an array of char pointers 100 large.
    // This is identical to passing `&a`, the address of `a`.
    function(aPtr);
    // `a` in this context "decays" to a pointer to its first element. Its first element is
    // `char*`, so here `a` is `char**`. And this is the signature you'll find in function2
    function2(a);

    // print out in main to show all the assignments were made
    printf("in main\n");
    printf("%s\n",a[0]);
    printf("%s\n",a[1]);
    printf("%s\n",a[2]);

    return 0;
}

演示

uj5u.com熱心網友回復：

您需要傳遞陣列的指標，即

function(a)

標籤：数组 C 细绳 指针

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