[英]How to read a string array from input in C and pass to function?
I have been stuck on this all day.我一整天都被困在这个问题上。 I want to be able to create a string array in C, and pass that array to the PrintStuff function.
我希望能够在 C 中创建一个字符串数组,并将该数组传递给 PrintStuff 函数。 I don't want to change my parameters to PrintStuff, but still make it work.
我不想将我的参数更改为 PrintStuff,但仍要使其工作。 Any help please?
请问有什么帮助吗?
void PrintStuff(const char **arr) {
for (int i = 0; i < 5; ++i) {
printf ("%s\n", arr[i]);
}
}
int main ()
{
//This works
//char * array[5] = { "this", "is", "a", "test", "megan"};
//This doesn't work
char * array[5];
for (int i=0;i<5;i++)
{
//scanf("%9s", array[i]);
fgets(array[i], 10, stdin);
}
Sort(array, 0, 5 - 1);
}
It doesn't do anything and I get this warning that says它没有做任何事情,我收到这个警告说
passing 'char *[5]' to parameter of type 'const char **' discards qualifiers in nested pointer types [-Wincompatible-pointer-types-discards-qualifiers]
将“char *[5]”传递给“const char **”类型的参数会丢弃嵌套指针类型中的限定符 [-Wincompatible-pointer-types-discards-qualifiers]
I've got no ideas what that means or how to fix it, HELP ME PLEASE!!!!!!!我不知道这意味着什么或如何解决它,请帮助我!!!!!!!
First Understand what is char* array[5]
.首先了解什么是
char* array[5]
。
that is nothing but array of 5 pointers to char*
, so you need memory for each pointer before you can use them.这只不过是指向
char*
的 5 个指针的数组,因此在使用它们之前,每个指针都需要内存。
char* array[5] = NULL;
int noe = sizeof(array) / sizeof (array[0]);
for(int a = 0; a < noe; a++) {
if( !(array[a] = malloc(BUFFER_SIZE))
printf("malloc failed at %d entry\n", a);
}
dont forget to free them when you are done using them使用完后不要忘记释放它们
for(int a = 0; a < noe; a++) {
if(array[a]) {
free(array[a]);
array[a] = NULL;
}
}
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