format '%s' expects argument of type 'char ', but argument 3 has type 'char ()[10]'

分类: 文章 • 2024-04-15 21:42:36

format '%s' expects argument of type 'char *', but argument 3 has type 'char (*)[10]'

实现 char name[][10];

You can just pass name by itself, which will evaluate to a pointer to the first element of the array( the same as &name[i] )

format '%s' expects argument of type 'char *', but argument 3 has type 'char (*)[10]'