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Question

#include <stdio.h> 
int bitCount(unsigned int n); 

int main(void) { 
    printf ("# 1-bits in base 2 representation of %u = %d, should be 0\n", 0, bitCount (0)); 
    printf ("# 1-bits in base 2 representation of %u = %d, should be 1\n", 1, bitCount (1)); 
    printf ("# 1-bits in base 2 representation of %u = %d, should be 17\n", 2863377066u, bitCount(2863377066u)); 
    printf ("# 1-bits in base 2 representation of %u = %d, should be 1\n", 268435456, bitCount(268435456)); 
    printf ("# 1-bits in base 2 representation of %u = %d, should be 31\n", 4294705151u, bitCount(4294705151u)); 
    return 0; 
} 

int bitCount(unsigned int n) { 
    /* your code here */ 
} 

你已经决定要你上面的位计数程序从命令行

# ./bitcount 17 
2 
# ./bitcount 255 
8 
# ./bitcount 10 20 
too many arguments! 
# ./bitcount 
[the same result as from part a] 

我得到，我们必须包括以上低于return 0 printf("too many arguments!")工作，但它不断给我一个错误。 任何人都可以帮助我吗？

Answer 1

修改您main接受申报参数：

int main(int argc, char * argv[]) { 

确保参数计数（argc）是2（一个用于指令，一个用于参数）：

if(argc < 2) { 
    // Give some usage thing 
    puts("Usage: bitcount <whatever>"); 
    return 0; 
} 

if(argc > 2) { 
    puts("Too many arguments!"); 
    return 0; 
} 

然后，使用类似atoi的东西将参数argv[1]解析为int：

printf("%d\n", bitCount(atoi(argv[1]))); 

（这在stdlib.h，顺便说一下。 ）