c++程序填空字符串获取字符

Question

获取字符串的前n个字符（默认的函数参数）
#include
#include
//获取字符串的前n个字符,n默认值为1
char * left(______________________________) ;
int main() {
char s[]="fdsfdsfds234234543543scfxzDSGFD*&ASGFwR12%$^";
std::cout<<left(s)<<':'<<left(s,10)<<':'<<left(s,100)<<std::endl;
return 0;
}
char *left(const char *s,int n) {

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}
输入
无
输出
字符串"fdsfdsfds234234543543scfxzDSGFD*&ASGFwR12%$^"的前1、10、100个字符。

Answer 1

 char * left(char * s, int n = 1) 
 {
 char * p = (char *)malloc(sizeof(char) * (n + 1));
 for (int i = 0; i < n; i++) p[i] = s[i];
 p[n] = '\0';
 return p;
}

问题解决的话，请点下采纳