Question

关于 vc++ char* 的问题

int a=4;
char * b[]={"1","2","3","4","5"};
现在我想要把a转化成字符形并且赋值给b[]的第一个字符串,使其变成{"4","2","3","4","5"},应该怎么做呢?
我试过这么做,但是报错了:
_itoa(a,b[0],10);

Answer 1

char * b[]={"1","2","3","4","5"};
相当于 char * b[5]={"1","2","3","4","5"};
就是定义5个指针，分别指向5个字符串"1","2","3","4","5",
我们知道这个五个字符串是常量，放在程序的静态缓冲区，是只读的。不可以修改只读的内存区。

就像 char*p = "hello word"; p[0] = 'H';//这样做是非法的

你的代码可以改成:

char a[5][2]={"1","2","3","4","5"};//不是静态内存
char * b[]={a[0],a[1],a[2],a[3],a[4]};
int a=4;
b[0][0] = '0'+a;

Answer 2

最好用
char b[5][10] = {"1","2","3","4","5"};
itoa(a,*(char*(&b[0])),10);
或者
char b[5] = {'1','2','3','4','5'};
itoa(a,b[0],10);

Answer 3

试一下，
CString s;
s.Format(" %s ",a);
再把s赋值进去

Answer 4

不能用指针数组，如果你这么赋初值就相当于让指针指向"4","2","3"这样的字符串常量，常量当然不允许修改了。
只能用字符数组解决这个问题。