Solve Exercise 7-1: `upper` and `lower`

Two cases, one file. This exercise was a welcome departure from the
recent head-scratchers. It made me think about a solution for the
pathname issue I realized would crop up, too.

I'm roughly 75% done with the book! Maybe by the time I finish it, I'll
have a Makefile ready for the project...

1 files changed, 52 insertions, 0 deletions

diff --git a/ch7/7-01_upper-lower.c b/ch7/7-01_upper-lower.c
new file mode 100644
index 0000000..2f641c1
--- /dev/null
+++ b/ch7/7-01_upper-lower.c
@@ -0,0 +1,52 @@
+#include <stdio.h>
+#include <ctype.h>
+#include <string.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 7-1: Write a program that converts upper case to lower or lower
+ * case to upper, depending on the name it is invoked with, as found in
+ * `argv[0]`.
+ *
+ * Notes: This one is, strangely, very easy. The primary gotcha is in the
+ * argv[0] check. The simplest solution I came up with was to check argv[0]
+ * for *both* upper and lower. If for some reason they're both in the path,
+ * the one that occurred later in the string takes precedence, since it's
+ * most likely to be the name of the executable itself rather than the
+ * directory containing the binary.
+ *
+ * Edge cases, gotta love'em.
+ */
+
+int main(int argc, char *argv[]) {
+	char c;
+	int lpos, upos;
+	char *plow, *pupp;
+	while ((c = getchar()) != EOF) {
+		plow = strstr(argv[0], "lower");
+		pupp = strstr(argv[0], "upper");
+		if (plow != NULL) {
+			lpos = (int) (plow - argv[0]);
+		} else {
+			lpos = -1;
+		}
+		if (pupp != NULL) {
+			upos = (int) (pupp - argv[0]);
+		} else {
+			upos = -1;
+		}
+		if (isalpha(c)) {
+			/* These two are only equal when neither are found in argv[0].
+			 * The odds of this happening are slim, but still possible.
+			 */
+			if (lpos > upos) {
+				c = tolower(c);
+			}
+			if (upos > lpos) {
+				c = toupper(c);
+			}
+		}
+		putchar(c);
+	}
+	return 0;
+}