CODEWITHSUNDEEP
c
Anirudh Ramanathan
Anirudh Ramanathan

Reputation: 46778

Passing char-array by value when char expected

void func(char a, int num)
{
    printf("%c",a);
}

int main()
{
     func("a", 6); //not func('a',6);
     printf("\n");
     func("b", 2); //not func('b',6);
}

I understand I am passing a char array of a and b with a null character \0. Could someone how it ends up printing the characters $ and &?

Upvotes: 1

Views: 257

Answers (2)

perreal
perreal

Reputation: 98118

You are passing pointer to a literal string but the func expects a character. Change it to receive an array:

void func(char *a, int num)
{
    printf("%c",a[0]); // also note that to print a char you need to
                       // 'select' a char from the array
}

Otherwise you'll end up printing to a char that is the ascii representation of whatever the address of a has as the first byte.

Upvotes: 2

MByD
MByD

Reputation: 137432

It could end up printing anything pretty much, probably part of the adresses of "a" and "b" match the ascii code of $ and &.

Upvotes: 4

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