如何正确分配新的字符串值?我试图理解如何以最干净/最安全的方式用C来解决这个琐碎的问题。下面是我的例子:#include <stdio.h>int main(int argc, char *argv[]){
typedef struct
{
char name[20];
char surname[20];
int unsigned age;
} person;
//Here i can pass strings as values...how does it works?
person p = {"John", "Doe",30};
printf("Name: %s; Age: %d\n",p.name,p.age);
// This works as expected...
p.age = 25;
//...but the same approach doesn't work with a string
p.name = "Jane";
printf("Name: %s; Age: %d\n",p.name,p.age);
return 1;}编译器的错误是:在函数‘main’中:main.c:18:错误:当从类型‘char*’分配给类型‘char[20]’时,不兼容类型我知道C(不是C+)没有字符串类型,而是使用字符数组,因此另一种方法是修改示例结构以保存字符的指针:#include <stdio.h>int main(int argc, char *argv[]){
typedef struct
{
char *name;
char *surname;
int unsigned age;
} person;
person p = {"John", "Doe",30};
printf("Name: %s; Age: %d\n",p.name,p.age);
p.age = 25;
p.name = "Jane";
printf("Name: %s; Age: %d\n",p.name,p.age);
return 1;}这如预期的那样有效,但我不知道是否有更好的方法来做到这一点。谢谢。
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![?](http://img1.sycdn.imooc.com/533e4c9c0001975102200220-100-100.jpg)
跃然一笑
TA贡献1826条经验 获得超6个赞
strncpy()
strncpy(p.name, "Jane", 19);p.name[19] = '\0'; //add null terminator just in case
strncat()
memcpy()
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