If I have a char *null_terminated_array="hello", I think printf("%zu",strlen(null_terminated_array)) should print 5 (not including '\0') and not 6 (including '\0').

CodePudding user response：

Yes that is what you would expect. Just be cautious if you were tempted to use sizeof rather than strlen on the array:

char *null_terminated_array="hello";
char char_array[6]="hello";    

printf("null_terminated_array = %s\n",null_terminated_array);
printf("strlen(null_terminated_array) = %zu\n",strlen(null_terminated_array));
printf("sizeof(null_terminated_array) = %zu\n",sizeof(null_terminated_array));
printf("\n");
printf("char_array = %s\n",char_array);
printf("strlen(char_array) = %zu\n",strlen(char_array));
printf("sizeof(char_array) = %zu\n",sizeof(char_array));

for the platform that I ran this on this gives

null_terminated_array = hello
strlen(null_terminated_array) = 5
sizeof(null_terminated_array) = 8

char_array = hello
strlen(char_array) = 5
sizeof(char_array) = 6

The apparent discrepancy for sizeof(null_terminated_array) is to do with automatic memory allocation using blocks of 32bits (platform dependent).