CODEWITHSUNDEEP
c
user1379351
user1379351

Reputation: 785

Why does changing the character of a string not give the desired character?

I am trying to edit a string after initialization using character literals, as follows:

int main() {
    char str1[10] = "hello";
    str1[0] = "b";
    printf("%s\n", str1);
    return 0;
}

The result is "dello" i.e. a "d" not a "b". Likewise the below just gives nonsense.

int main() {

    char str1[10];

    str1[0] = "h";
    str1[1] = "e";
    str1[2] = "l";
    str1[3] = "l";
    str1[4] = "o";

    printf("%s\n", str1);
}

I found one StackOverflow post that mentioned that this should cause a segfault or an Access-Violation error, but it doesn't really explain why.

Upvotes: 0

Views: 99

Answers (2)

Emil Laine
Emil Laine

Reputation: 42838

str1[0] = "b";

Here, "b" is a string literal, not a character. Characters are enclosed in single quotes:

str1[0] = 'b';

If you had compiler warnings enabled you'd get something like:

warning: incompatible pointer to integer conversion assigning to 'char' from 'char [2]' [-Wint-conversion]

str1[0] = "b";
        ^ ~~~

In your second code, your string is missing a terminating null-character, and so passing it to printf invokes undefined behavior because printf can't know where your string ends. To append the null-character at the end, just do:

str1[5] = '\0';

Upvotes: 12

Pavel B
Pavel B

Reputation: 106

In C, single quotes identify chars and double quotes create a string literal.

Try doing the following: 

str1[0] = 'b'; //note the single quote

Upvotes: 3

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