How to initialize string in struct without using strcpy function?

Question

Is there any manual way to initialize the string in struct ? I used to initialize string in struct using strcpy function such as:

typedef struct {
    int id;
    char name[20];
    int age;
} employee;


int main()
{
    employee x;
    x.age=25;

    strcpy(x.name,"sam");
    printf("employee age is %d \n",x.age);
    printf("employee name is %s",x.name);

    return 0;
}

to copy into a fixed-size array, you should always use strncpy(), or strncpy_s, if available. — Michaël Roy
– Michaël Roy, Commented Aug 20, 2017 at 0:12
@MichaëlRoy Disagree with "to copy into a fixed-size array, you should always use strncpy(), ..." — chux
– chux, Commented Aug 20, 2017 at 3:29
You can. But don't come posting questions here about buffer overflow. — Michaël Roy
– Michaël Roy, Commented Aug 20, 2017 at 3:31

Vlad from Moscow · Accepted Answer · 2017-08-20 00:20:44Z

7

Strictly speaking this

strcpy(x.name,"sam");

is not an initialization.

if to speak about the initialization then you can do it the following way

employee x = { .name = "sam", .age = 25 };

or

employee x = { .name = { "sam" }, .age = 25 };

This is equivalent to the following initialization

employee x = { 0, "sam", 25 };

or

employee x = { 0, { "sam" }, 25 };

Or you even can use a compound literal of the type employee to initialize the object x though that is not efficient.

Otherwise if is not an initialization but an assignment of the data member of the structure then indeed you have to use at least strcpy or strncpy.

edited Aug 20, 2017 at 0:20

answered Aug 19, 2017 at 23:46

Vlad from Moscow

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2 Comments

chux Over a year ago

Good answer aside from "compound literal ... though that is not efficient.". Certainly a good compiler will optimize that as well as other approaches.

chux Over a year ago

Assignment can use a compound literal without strcpy() and friends.

0___________ · Accepted Answer · 2017-08-20 08:36:12Z

-1

max - including trailing zero

char *mystrncpy(char *dest, const char *src, size_t max)
{
    char *tmp = dest;

    if (max)
    {
        while (--max && *src)
        {
            *dest++ = *src++;
        }

    *dest++ = '\0';
    }
    return tmp;
}

edited Aug 20, 2017 at 8:36

answered Aug 19, 2017 at 23:59

0___________

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1 Comment

chux Over a year ago

*dest++ = '\0'; belongs inside if(max)

lost_in_the_source · Accepted Answer · 2017-08-20 12:25:26Z

-1

You can write your own version of strcpy:

void mycopy(char *dest, const char *source, size_t ndest)
{
    assert(ndest != 0);
    while (--ndest > 0 && (*dest++ = *source++))
        ;
}

You're not using strcpy anymore. Plus it is safer.

edited Aug 20, 2017 at 12:25

answered Aug 19, 2017 at 23:37

lost_in_the_source

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10 Comments

gnasher729 Over a year ago

You should have clearly marked this as "sarcastic". I assume it is sarcastic. If not, then your employer should be frightened.

0___________ Over a year ago

I have not seen so badly written simple function for years

0___________ Over a year ago

going to bed enough for today

ad absurdum Over a year ago

There need to be parentheses around the assignment since && has higher precedence than =.

chux Over a year ago

Corner case weakness: mycopy(dest, source, 0) leads to buffer overrun. strncpy() does not have that problem.

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