I wrote a small snippet of C code. Ignore the strange "sentence" name. My question is, why is it that when I call strlen on this array, it prints 5, instead of 2? Thanks.
Code:int main(){ char sentence [2]; sentence[0]= 'y'; sentence[1]='o'; int len= strlen(sentence); printf("%d", len); }
Because you didn't leave room for the null terminator, leading to a buffer overflow.
Code:int main(){ char sentence [3]; sentence[0]= 'y'; sentence[1]= 'o'; sentence[2]= 0; int len= strlen(sentence); printf("%d", len); return 0; }
Strings end with a null byte. Your code needs to be changed to:
Code:int main(void){ char sentence [3]; size_t len; sentence[0] = 'y'; sentence[1] = 'o'; sentence[2] = '\0'; len= strlen(sentence); printf("%u", len); return 0; }
bit∙hub [bit-huhb] n. A source and destination for information.
Before fixing it, change your program to this to see what you get:
You'll note that storage is not initialized, unless you specifically tell the compiler (or runtime) to initialize it.Code:int main(){ unsigned char sentence [2]; printf("1: sen[0]=%01x, sen[1]=%01x, sen[2]=%01x\n", sentence[0], sentence[1], sentence[2]) ; sentence[0]= 'y'; sentence[1]='o'; printf("2: sen[0]=%02x, sen[1]=%02x, sen[2]=%02x\n", sentence[0], sentence[1], sentence[2]) ; int len= strlen(sentence); printf("%d", len); }
Mainframe assembler programmer by trade. C coder when I can.
use %zu to print size_t, or %lu and cast to unsigned long in C89.
