7月8号作业
2020-07-14 06:33
127 查看
void Print1(char* p)
{
printf(“p = %s\n”,p);
}
void Print2(char(*k)[100])
{
int i = 0;
for (; i < 2; i++)
{
printf(“k[%d]=%s\n”, i, *(k + i));
}
}
void Print3(char(*s)[2][100])
{
int i = 0; int j = 0;
for (; i < 2; i++)
{
for (; j < 2; j++)
{
printf(“s[%d][%d]=%s\n”,i+1,j, (s)[i][j]);
}
}
}
int main()
{
char p[100] = “hello1”;
char k[2][100] = { “hello word”, “hello word hello” };
char s[2][2][100] = { {“hello2”, “hello4”},{“hello5”, “hello6”} };
Print1§;
Print2(k);
Print3(s);
return 0;
}
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