通过字符串迭代C中的分段错误

我对C非常陌生，我应该创建strtok函数。问题是，我不断收到分段错误在此代码：通过字符串迭代C中的分段错误

char *newstring = "testing, testing, tested." 
while(*newstring != '\0'){ 
    /*replace delimiters with nulls*/ 
    printf("STARTING ITERATION\n"); 
    if(*newstring ==','){ 
     *newstring='\0'; 
    } 

    printf("newstring char in loop: %c\n", *newstring); 
    /*printf("delimiters char in loop: %c\n", *delimiters);*/ 
    newstring++; 
    printf("END OF ITERATION\n"); 
} 
printf("OUT OF ITERATION");

输出是这样的：

STARTING ITERATION 
newstring char in loop: t 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: e 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: s 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: t 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: t 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: e 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: s 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: t 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: t 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: e 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: s 
END OF ITERATION 
STARTING ITERATION 
newstring char in loop: . 
END OF ITERATION 
Segmentation fault (core dumped)

所以，它看起来像它会一路走过来的字符串的结束和循环的最后一步，但实际上并没有使它退出循环。我不知道我错过了什么。

来源

2016-09-19 ChasetopherB

你正在修改字符串字面..这是未定义的行为。试试'char newstring [] =“测试，测试，测试。”;'而是。 – yano

指针不是数组。您的代码调用未定义的行为。您尝试修改_string literal_。 – Olaf

你知道Literal String是什么吗？ – Michi