为什么我的 bash 脚本被“杀死”了？我该如何防止？

Question

我编写了一个 bash 脚本来针对测试套件测试学生编写的 C 程序。由于某种原因，脚本在一段时间后被杀死。我是 bash 脚本的新手，直到现在还没有找到原因。这是脚本。

#!/bin/bash
ulimit -t 1
tests_dir=tests
run_dir=tests
find . -name "*.c" | while read cfile; do rm a.out &> /dev/null; gcc "$cfile" -lm -w &> /dev/null;
if [ ! -f a.out ]; 
then 
    echo "$cfile did-not-compile" >> "$run_dir/results.out";
else
    find "$tests_dir" -name "in*.txt" | while read testin; do echo "running $testin on $cfile"; 
    rm test.out &> /dev/null; 
    rm space_less_testout &> /dev/null;
    LD_PRELOAD=../../EasySandbox/EasySandbox.so ./a.out < $testin | grep -v "entering SECCOMP mode" &> test.out;
    if [ -e test.out ]; then
            testout=${testin/in/out}
            tr -d '\n' < $testout | tr -d ' ' > space_less_testout
            echo -e '\n' >> space_less_testout

            if diff -qwB "$testout" test.out &> /dev/null
            then
                    # if no difference then takes true brance (based on return value)
                    echo "$cfile ;passed-on-test; $testin" >> "$run_dir/results.out"; echo "passed-on-test $testin";
            elif diff -qB space_less_testout test.out &> /dev/null 
            then
                    # or no difference with new-line removed should-be-output (just a formatting error)
                    echo "$cfile ;passed-on-test; $testin" >> "$run_dir/results.out"; echo "passed-on-test $testin";
            else
                    echo "$cfile ;failed-on-test; $testin" >> "$run_dir/results.out"; echo "failed-on-test $testin";
            fi
    fi
done;
fi
done;

Answer 1

ulimit -t 1将脚本的 CPU 时间限制为 1 秒。当脚本耗尽其所有 CPU 时间时，它就会被杀死。

要限制脚本中仅一个命令的 CPU 时间，您可以使用括号在具有自己限制的子 shell 中启动它，例如

(ulimit -t 1; LD_PRELOAD=../../EasySandbox/EasySandbox.so ./a.out < $testin)