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authorzlg <zlg@zlg.space>2013-02-27 05:54:58 -0600
committerzlg <zlg@zlg.space>2013-02-27 05:54:58 -0600
commit8557e29d33b365816114996d04531468e13e78b1 (patch)
tree9206150286154c02bb804734151845001cd1767d
parentShorten 2-01's code (diff)
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Solve Exercise 2-02: No logical operators
The text doesn't specify which technique to use. The section outlined operator precedence, but I couldn't think of a way to check for truth in the three main expressions without logical operators. So I turned to a recursive function. It works, but I'm not sure if it's what K&R were after.
-rw-r--r--ch2/2-02_no-logical-operators.c44
1 files changed, 44 insertions, 0 deletions
diff --git a/ch2/2-02_no-logical-operators.c b/ch2/2-02_no-logical-operators.c
new file mode 100644
index 0000000..6fea5d6
--- /dev/null
+++ b/ch2/2-02_no-logical-operators.c
@@ -0,0 +1,44 @@
+#include <stdio.h>
+
+/* The C Programming Language, 2nd Edition
+ *
+ * Exercise 2-2: Write a loop equivalent to the 'for' loop above without
+ * using && or ||.
+ *
+ * (The for loop is:
+ *
+ * for (i=0; i < lim - 1 && (c = getchar()) != '\n' && c != EOF)
+ * i++;
+ *
+ * Answer: A recursive function should do the job. I think...
+ *
+ */
+
+#define LIMIT 17
+
+char test[LIMIT] = "";
+int i = 0;
+int c;
+
+void save_string(char s[]) {
+ if (i < LIMIT - 1) {
+ c = getchar();
+ if (c != '\n') {
+ if (c != EOF) {
+ s[i] = c;
+ i++;
+ } else {
+ return;
+ }
+ } else {
+ return;
+ }
+ save_string(s);
+ }
+}
+
+int main() {
+ save_string(test);
+ printf("%s\n", test);
+ return 0;
+}