如何用write()输出字符串指定索引字符？解决参数类型警告问题

Let's break down why your code is throwing that warning and failing to produce output, then fix it step by step.

The Root Problem #

The write() system call expects its second argument to be a pointer to the data you want to write (const void*), but you’re passing name[strlen(name)-1]—which is a single char value (the ASCII code for '3', which is 51). When you pass an integer where a pointer is required, the compiler treats that integer as a memory address. So write() tries to read 1 byte from address 0x33 (hex for 51), which is invalid memory—hence no output and the "pointer from integer without a cast" warning.

The Simple Fix: Pass the Character's Address #

Instead of passing the character itself, pass its memory address using the & (address-of) operator. This gives write() the valid pointer it needs:

#include <unistd.h>
#include <string.h>

int main() {
    char name[] = "Name123";
    size_t str_length = strlen(name);
    
    // Add a safety check to avoid index underflow if the string is empty
    if (str_length > 0) {
        // Pass the address of the last character
        write(2, &name[str_length - 1], 1);
    }
    return 0;
}

What’s Different Here? #

• &name[str_length - 1] returns a pointer directly to the '3' character in the array, which matches the const void* type write() expects.
• We added a check for str_length > 0 because strlen() returns a size_t (unsigned integer). If the string were empty, str_length - 1 would wrap around to a huge value, causing an out-of-bounds access.

Why This Works #

When you run this corrected code, write() reads 1 byte from the valid address of the '3' character and writes it to file descriptor 2 (stderr, which prints directly to your terminal). You’ll see the desired '3' output as expected.

内容的提问来源于stack exchange，提问作者Junaid