So I was wondering why I can't do this.
int main(int argc, char **argv){
FILE *src;
char filePath[261];
filePath = argv[1];
The last line is where there is a compiler error. What's the difference between char[] and char*? How would I fix this code so I can set filePath equal to argv[1]. Thanks in advance.
Use
strcpy(filePath, argv[1]);
and live happy. Don't forget to check argv[1] for being NULL and don't forget to see if argc is > 1.
Your filePath variable is a fixed-size array which is allocated on the stack and argv[i] is a pointer to some memory in the heap. Assigning to filePath cannot be done, because filePath is not a pointer, it is the data itself.
Because filePath is an array and it's not allowed to modify the address of an array in C.
You can use the string family to copy the strings.
You need to have:
FILE *src;
char *filePath;
filePath = argv[1];
since filePath must point to argv, not to an array of 261 bytes. If you want, you can copy the argumenti into the array:
FILE *src;
char filePath[261];
strcpy(filePath, argv[1]);
or better, to avoid risking copying more bytes than you have available (which would result in disaster):
FILE *src;
char filePath[261];
strncpy(filePath, argv[1], sizeof(filePath));
or again
#define MAX_FILESIZE 261
FILE *src;
char filePath[MAX_FILESIZE];
strncpy(filePath, argv[1], MAX_FILESIZE);
问：char[] 和 char* 有什么区别？
A：很多时候，你可以互换使用它们。
但是在这里，您“尝试将数组名称用作左值”:)
这是一个很好的解释：
以下是“什么是合法的，什么不是”的简短摘要：