Why doesn't this work.
printf("%s\n", argv[1][3]);
When this works?
printf("%c\n", argv[1][3]);
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2 Answers
Because the %s format specifier tells printf that the argument is a null-terminated string. You're giving printf a single character--the fourth character in the second element of the argv array.
If you want to print the substring from the fourth character to the end of the string, you can do that too, you just have to get a pointer to that character:
printf("%s\n", &argv[1][3]);
or, if you prefer:
printf("%s\n", argv[1] + 3);
"%s" in a foramt string expects a 'char *' argument, but you're passing it a 'char' so you get garbage (probably a crash). "%c" in a format string expects a 'char' argument, which is what you are giving it, so it works.
1 Comment
Chris Lutz
Less likely a crash than a compiler error. GCC (at least) can do type checking for format-string functions like
printf and scanf.