虽然尝试使用atoi函数将字符串转换为整数,但没有任何输出。调试后,它在t=atoi(s[i]);
行中显示分段错误错误这是供您参考的代码:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main()
{
char s[100];
int i=0,t;
printf("Enter: ");
fgets(s,100,stdin);
while(s[i]!='\0')
{
if(s[i]>='1' && s[i]<='9')
{
t = atoi(s[i]);
printf("%d\n",t);
}
i++;
}
return 0;
}
编译时,请始终启用警告,然后修复这些警告:
通过gcc
编译器运行发布的代码将导致:
gcc -O1 -ggdb -Wall -Wextra -Wconversion -pedantic -std=gnu11 -c "untitled2.c" -I. (in directory: /home/richard/Documents/forum)
untitled2.c: In function ‘main’:
untitled2.c:14:16: warning: passing argument 1 of ‘atoi’ makes pointer from integer without a cast [-Wint-conversion]
t = atoi(s[i]);
^
In file included from /usr/include/features.h:424:0,
from /usr/include/x86_64-linux-gnu/bits/libc-header-start.h:33,
from /usr/include/stdio.h:27,
from untitled2.c:1:
/usr/include/stdlib.h:361:1: note: expected ‘const char *’ but argument is of type ‘char’
__NTH (atoi (const char *__nptr))
^
untitled2.c:9:3: warning: ignoring return value of ‘fgets’, declared with attribute warn_unused_result [-Wunused-result]
fgets(s,100,stdin);
^~~~~~~~~~~~~~~~~~
Compilation finished successfully.
换句话说,此语句:
t = atoi(s[i]);
正在向函数传递单个字符:atoi()
但是,atoi()
希望传递给char数组的指针。
在atoi()
的MAN页面中,语法为:
int atoi(const char *nptr);
建议:替换:
t = atoi(s[i]);
printf("%d\n",t);
with:
printf( "%d\n", s[i] );
将输出数组s[]
中每个字符的ASCII值。例如,ASCII值“ 1”为49。
请注意,现代编译器输出的警告是无法检查C库函数的返回值。