我正在练习字符串数组，但好像不理解它如何工作，有人能向我解释一下正确的编写代码方式吗？
代码目标：将两个字符串赋值给一个字符串数组，并使用函数打印它们。
错误：编译器没有发现任何错误，但是我在终端中没有得到任何输出。
#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]);
}
function(&a[MAX]);
你正在传递数组 a 中索引为 MAX 的项的地址，但是你的函数将其解释为整个数组的地址。你应该改为写成：
function(a)
char *s[]来表示字符指针数组。但是在调用函数时，编译器已经知道函数需要的类型，所以您只需要传递与声明类型匹配的变量即可。在这种情况下，像a[MAX]这样的方括号被解释为选择数组的一个元素的下标。function是一个糟糕的函数名称。我知道这只是一个小型测试程序，所以没关系，但请尝试养成给事物命名具有描述性的习惯。printf("%s",*a[1]); printf("%s",*a[0]);
在这里，您正在访问索引为1的项目，但是您还没有在那里存储任何内容。这不是一个好计划。要么删除第一行，要么更改代码以在索引1中存储某些内容。
此外，您不需要取消引用元素。数组中的每个元素都是字符数组，printf()将期望char *类型的值，这就是每个a[n]将成为的内容。所以只需使用：
printf("%s", a[1]);
printf("%s", a[0]);
a[1] 了。无论如何，是的，将数组称为 a 提供了数组的值，实际上只是第一个元素（索引为0的元素）的地址。出于同样的原因，您的 printf() 语句应该将字符串称为例如 a[0]，而不是 *a[0]。*a[0] 是数组中第一个字符串的第一个字符。请注意，我刚才添加了这一点的编辑。 - Calebchar *s = a [0]; char c = s [2]; - Caleb你似乎对你正在使用的指针及其正确类型存在一些困惑。希望下面的程序可以澄清一些事情。注释中有解释：
#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;
}
a[0][2]。在 function2 中，a 是一个 char**，所以 a[0] 是第一个 char*，而 a[0][2] 是从第一个 char* 开始偏移的第三个 char。你可以去 Demo 链接并在那个沙盒中玩耍，不要忘记 "%c" 是打印 char 的格式说明符。 - yanofunction(a)
"%d"吗？出了什么问题？ - doctorlovea中最后一个元素之后的地址，这也是错误的类型。你应该在这里得到一堆警告。 - Eugene Sh.