char x[3][3][3]={{"bad","had","rad"},{"bat","cat","hat"},{"hit","git","bit"}};
printf("%s\n", x[1][0]);
Why is the output for this code "batcathathitgitbit" and not just "bat"? I am having difficulty wrapping my head around why it is not just printing one item.
Strings must have a null terminator in C, it tells the compiler where to stop parsing for the string. Make every string set one longer so that there is still a null character at the end
char x[3][3][4]={{"bad","had","rad"},{"bat","cat","hat"},{"hit","git","bit"}};
printf("%s\n", x[1][0]);
Well, think about the type your printf is taking. You told it you'll pass an %s, i.e. a nul-terminated string, with a const char* argument.
Now think about your 3-dimensional array. How is it actually laid out in memory? As the characters of consecutive cells:
x[0][0][0] - that's 'b',
x[0][0][0] - that's 'a',
x[0][0][2] - that's 'd',
... what's next? it's not x[0][0][3], since x's innermost dimension is 3, not 4! So we have no place to put the trailing '\0'. In fact, we now have:
x[0][1][0] - that's 'h',
x[0][1][1] - that's 'a',
etc.
Thus, only the first 3 characters of each string are used, and you get an uninterrupted sequence of characters; the string doesn't even end with the last character in x.
Make the innermost dimension large enough to fit all your words: char x[3][3][4]. Then you'll get the expected output (GodBolt).
Simple test program unveiles the memory representation of the 3d array:
int main(void)
{
char x[3][3][3]={{"bad","had","rad"},{"bat","cat","hat"},{"hit","git","bit"}};
char *ptr = (char *)x;
for(size_t i = 0; i < sizeof(x); i++)
{
printf("x[%2zu] = '%c'\n", i, ptr[i]);
}
}
x[ 0] = 'b'
x[ 1] = 'a'
x[ 2] = 'd'
x[ 3] = 'h'
x[ 4] = 'a'
x[ 5] = 'd'
x[ 6] = 'r'
x[ 7] = 'a'
x[ 8] = 'd'
x[ 9] = 'b'
x[10] = 'a'
x[11] = 't'
x[12] = 'c'
x[13] = 'a'
x[14] = 't'
x[15] = 'h'
x[16] = 'a'
x[17] = 't'
x[18] = 'h'
x[19] = 'i'
x[20] = 't'
x[21] = 'g'
x[22] = 'i'
x[23] = 't'
x[24] = 'b'
x[25] = 'i'
x[26] = 't'
