[英]Storing a char in an array segfault
I'm getting a segfault in my code and I'm not sure why. 我的代码中出现了段错误,我不确定原因。 I read through a file and count the number of lines in order to dynamically allocate my arrays.
我读了一个文件并计算了行数,以便动态分配我的数组。 I then rewind the file, read the data in the file, storing the data into variables, and then storing the read variables into arrays, but I'm having trouble with chars.
然后我回放文件,读取文件中的数据,将数据存储到变量中,然后将读取的变量存储到数组中,但是我遇到了chars问题。
...
char *aname = malloc(sizeof(char) * 3);
...
// get # lines in file (count)
...
char *aname_seen = malloc(count * (sizeof(char) * 3));
...
rewind(file);
while (fgets(buff, sizeof buff, file) != NULL)
{
if (sscanf(buff, "%s %d %s %s %d %lf %lf %lf %lf %lf\n",
atm, &serial, aname, resName, &resSeq, &x, &y, &z,
&occupancy, &tempFactor) == 10)
{
aname_seen[i] = *aname;
printf("%d: %s vs %s\n", i, aname, aname_seen[i]);
i++;
} // end sscanf if-loop
} // end while loop
I can print aname with printf("%d: %s\\n", i, aname)
and get the expected output, but I'm getting Segmentation fault (core dumped)
when I try printf("%d: %s vs %s\\n", i, aname, aname_seen[i])
. 我可以使用
printf("%d: %s\\n", i, aname)
打印aname并获得预期的输出,但是当我尝试printf("%d: %s vs %s\\n", i, aname, aname_seen[i])
时,我遇到了Segmentation fault (core dumped)
printf("%d: %s vs %s\\n", i, aname, aname_seen[i])
。
This while loop + nested if loop is the same convention I use to count the number of lines, so i
will increment up to count. 这个while循环+嵌套if循环是我用来计算行数的相同约定,所以
i
会递增计数。 Am I incorrectly allocating aname_seen and not actually giving it count
number of char*3
elements? 我是否错误地分配aname_seen并且实际上没有给它
count
char*3
元素的数量? I'm not well versed in messing with char's. 我不太擅长搞乱char。 More of a numerical fortran buff, so I need some direction.
更多的数字fortran buff,所以我需要一些方向。
Thanks in advance! 提前致谢!
The %s
format specifier is supposed to correspond to a char *
argument. %s
格式说明符应该对应于char *
参数。 In your case, aname_seen[i]
is a char
, which gets promoted to an int
for the purpose of passing to a variadic function ( printf
). 在你的例子中,
aname_seen[i]
是一个char
,它被提升为一个int
,目的是传递给一个可变函数( printf
)。 An int
is not a char *
. int
不是char *
。
Perhaps you meant one of these: 也许你的意思是其中之一:
printf("%d: %s vs %c\n", i, aname, aname_seen[i]);
printf("%d: %s vs %s\n", i, aname, &aname_seen[i]);
If neither of these solve your problem, please explain precisely the behaviour you expect from this expression and give us a minimal, compilable testcase. 如果这些都没有解决您的问题,请准确解释您对此表达式的期望,并为我们提供一个最小的,可编译的测试用例。 Your current testcase isn't compilable.
您当前的测试用例不可编译。
you way you defined aname_seen
is a pointer to a char array 你定义的方式
aname_seen
是一个指向char数组的指针
char *aname_seen = malloc(count * (sizeof(char) * 3));
so aname_seen[i]
is a char
所以
aname_seen[i]
是一个char
so the 所以
printf("%d: %s vs %s\n", i, aname, aname_seen[i]);
should be 应该
printf("%d: %s vs %c\n", i, aname, aname_seen[i]);
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