如何忽略'\ 40'表达式并输出单元格文本

Question

所以我总是使用figlet {text}为我的程序生成方框文本,但这次~My C编译器决定采取行动.如何忽略此文本数组的\ 40:

static char *header =
"  _   _      _                      _      _____          _              \n"
" | \ | | ___| |___      _____  _ __| | __ |_   _| __ ___ (_) __ _ _ __   \n"
" |  \| |/ _ \ __\ \ /\ / / _ \| '__| |/ /   | || '__/ _ \| |/ _` | '_ \  \n"
" | |\  |  __/ |_ \ V  V / (_) | |  |   <    | || | | (_) | | (_| | | | | \n"
" |_| \_|\___|\__| \_/\_/ \___/|_|  |_|\_\   |_||_|  \____/ |\__,_|_| |_| \n"
"                                                       |__/              \n";

如何在没有错误的情况下进行编译？

Answer 1

逃避所有反斜杠:

static const char header[] =
"  _   _      _                      _      _____          _              \n"
" | \\ | | ___| |___      _____  _ __| | __ |_   _| __ ___ (_) __ _ _ __   \n"
" |  \\| |/ _ \\ __\\ \\ /\\ / / _ \\| '__| |/ /   | || '__/ _ \\| |/ _` | '_ \\  \n"
" | |\\  |  __/ |_ \\ V  V / (_) | |  |   <    | || | | (_) | | (_| | | | | \n"
" |_| \\_|\\___|\\__| \\_/\\_/ \\___/|_|  |_|\\_\\   |_||_|  \\____/ |\\__,_|_| |_| \n"
"                                                       |__/              \n";

这在源代码中看起来很奇怪,但打印得很好.