通過字符串迭代C中的分段錯誤

我對C非常陌生，我應該創建strtok函數。問題是，我不斷收到分段錯誤在此代碼：通過字符串迭代C中的分段錯誤

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");

輸出是這樣的：

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，它看起來像它會一路走過來的字符串的結束和循環的最後一步，但實際上並沒有使它退出循環。我不知道我錯過了什麼。

來源

2016-09-19 ChasetopherB

你正在修改字符串字面..這是未定義的行爲。試試'char newstring [] =「測試，測試，測試。」;'而是。 – yano

指針不是數組。您的代碼調用未定義的行爲。您嘗試修改_string literal_。 – Olaf

你知道Literal String是什麼嗎？ – Michi