sed: fix commandline-given expression when -e is not used

Make explicit sed commands (first on commandline) behave the same
as those given with -e.

Without this patch the following two commands behave differently,
the second one being wrong:
echo ab | sed -e $'1 i\\\n--'
echo ab | sed $'1 i\\\n--'

Reviewed by:	0mp, des, kevans
Sponsored by:	Klara, Inc.

(cherry picked from commit 0552fdc62c)

usr.bin/sed/main.c

@@ -148,10 +148,8 @@ main(int argc, char *argv[])
 			break;
 		case 'e':
 			eflag = 1;
-			if ((temp_arg = malloc(strlen(optarg) + 2)) == NULL)
-				err(1, "malloc");
-			strcpy(temp_arg, optarg);
-			strcat(temp_arg, "\n");
+			if (asprintf(&temp_arg, "%s\n", optarg) == -1)
+				err(1, "asprintf");
 			add_compunit(CU_STRING, temp_arg);
 			break;
 		case 'f':
@@ -184,7 +182,9 @@ main(int argc, char *argv[])
 
 	/* First usage case; script is the first arg */
 	if (!eflag && !fflag && *argv) {
-		add_compunit(CU_STRING, *argv);
+		if (asprintf(&temp_arg, "%s\n", *argv) == -1)
+			err(1, "asprintf");
+		add_compunit(CU_STRING, temp_arg);
 		argv++;
 	}
 

usr.bin/sed/tests/sed2_test.sh

@@ -147,6 +147,18 @@ bracket_y_body()
 	    echo 'bra[ke]' | sed 'y[\[][ct['
 }
 
+atf_test_case minus_e
+minus_e_head()
+{
+	atf_set "descr" "Verify that -e and implicit arg do the same thing"
+}
+minus_e_body()
+{
+	printf "ab\n" > a
+	atf_check -o 'inline:--\nab\n' sed -e $'1 i\\\n--' a
+	atf_check -o 'inline:--\nab\n' sed    $'1 i\\\n--' a
+}
+
 atf_init_test_cases()
 {
 	atf_add_test_case inplace_command_q
@@ -156,4 +168,5 @@ atf_init_test_cases()
 	atf_add_test_case commands_on_stdin
 	atf_add_test_case hex_subst
 	atf_add_test_case bracket_y
+	atf_add_test_case minus_e
 }