c - 使用带字符串的长度子说明符的 printf 行为

Question

我很好奇 printf 是如何工作的，我尝试了很多技巧，比如:

printf("Test 1: %hs\n", "hi");
printf("Test 2: %hhs\n", "hi");
printf("Test 3: %ls\n", "hi");
printf("Test 4: %lls\n", "hi");
printf("Test 5: %js\n", "hi");
printf("Test 6: %zs\n", "hi");
printf("Test 7: %ts", "hi");

我将 printf 与 s 说明符和所有长度修饰符一起使用(通常与数字说明符一起使用(如 d/i/u/o ...)和 i得到奇怪的输出:

Test 1: hi
Test 2: hi
Test 3: Test 4: Test 5: Test 6: hi
Test 7: hi

似乎是 l/ll/j 长度修饰符导致 printf 错误并停止工作(它不打印 \n 但使用所有其他长度修饰符它似乎忽略了长度修饰符并像正常用法一样工作。

为什么会遵循这种行为？

Answer 1

答案是undefined behaviour .格式说明符 %s 不支持您用于传递的参数的任何长度修饰符。 C 中未定义格式说明符不匹配。当我使用 gcc 编译器 (gcc -Wall -Wextra -std=c11 test.c) 编译您的代码时，它表示:

test.c: In function ‘main’:
test.c:5:19: warning: use of ‘h’ length modifier with ‘s’ type character has either no effect or undefined behavior [-Wformat=]
 printf("Test 1: %hs\n", "hi");
                   ^
test.c:6:20: warning: use of ‘hh’ length modifier with ‘s’ type character has either no effect or undefined behavior [-Wformat=]
 printf("Test 2: %hhs\n", "hi");
                    ^
test.c:7:19: warning: format ‘%ls’ expects argument of type ‘wchar_t *’, but argument 2 has type ‘char *’ [-Wformat=]
 printf("Test 3: %ls\n", "hi");
                   ^
test.c:8:20: warning: use of ‘ll’ length modifier with ‘s’ type character has either no effect or undefined behavior [-Wformat=]
 printf("Test 4: %lls\n", "hi");
                    ^
test.c:9:19: warning: use of ‘j’ length modifier with ‘s’ type character has either no effect or undefined behavior [-Wformat=]
 printf("Test 5: %js\n", "hi");
                   ^
test.c:10:19: warning: use of ‘z’ length modifier with ‘s’ type character has either no effect or undefined behavior [-Wformat=]
 printf("Test 6: %zs\n", "hi");
                   ^
test.c:11:19: warning: use of ‘t’ length modifier with ‘s’ type character has either no effect or undefined behavior [-Wformat=]
 printf("Test 7: %ts", "hi");