Copying strings to an array

Question

In my code's output I find that the strings I am printing are all (null) even though I input count strings of characters < 31.

char name[31];
char *names[32];
int count = 5;
for (int i =0; i<count;i++) {
        scanf("%s",name);
        names[i]=(char*)malloc(strlen(name)+1);
        strcpy(names[i],name);
    }

for (int i =0; i<count;i++) {
    printf("%d: %s\n",i+1,names[count]);
}

Probably a typo, but printf("%d: %s\n",i+1,names[count]); should be printf("%d: %s\n",i+1,names[i]); - otherwise, you're printing an unallocated string. — Adrian Mole
– Adrian Mole, Commented Oct 22, 2020 at 22:13

Chems Eddine · Accepted Answer · 2020-10-22 22:13:45Z

2

Maybe you should use names[i] in stead of names[count]

for (int i = 0; i < count; i++) {
    printf("%d: %s\n", i+1, names[i]);
}

answered Oct 22, 2020 at 22:13

Chems Eddine

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LeonFibuck · Accepted Answer · 2020-10-22 22:20:04Z

2

you're indexing names with count but your code initialize string only up to count - 1. Changing printf this way:

printf("%d: %s\n",i+1,names[i]);

should do the trick.

answered Oct 22, 2020 at 22:20

LeonFibuck

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Comments

Zig Razor · Accepted Answer · 2020-10-23 06:42:03Z

1

Your error is in the printf function, you need to modify it in this way:

printf("%d: %s\n",i+1,names[count]); => printf("%d: %s\n",i+1,names[i]);

edited Oct 23, 2020 at 6:42

Zig Razor

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answered Oct 22, 2020 at 22:13

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